LIBRARY OF CONGRESS 



029 942 845 A 



andardized Square Footings 

OF 

Reinforced Concrete 



Standardized Square Footings 

OF 

Reinforced Concrete 



By 

LEWIS A. HICKS 

Structural Engineer 
Rialto Building : : San Francisco 



Sehwabacher-Frey Stationery Co. 

Printers 

No 609-611 Market St., San Francisco 

1915 



^ 



^\4 



CONTENTS 



PAGE 

1. Introduction 1 

2. Plates 6 

3. Notation 4 

Working Stresses 9 

Formulae 10 

4. Use of Diagrams — 

a. Design — For Architects and Engineers 12 

b. Review — Building Departments and Checkers ... 20 

c. Experiment 28 

5. General Information — For Student Instruction .... 32 



/ 

Copyright 1915 
dy Lewis A. Hicks 

©01,4406642 
JUL -9 1915 

0/ 



/5~- ISbSS' 



INTRODUCTION. 



The text of this booklet is merely the explanation of the use of 
the diagrams, devised by the writer for the solution of the problems 
arising in connection with the design and review of Square Footings 
of Reinforced Concrete. 

The economic advantages offered by concrete footings reinforced 
with steel bars, as contrasted with the earlier form, using structural 
steel shapes, are so great, that the importance of correct design is 
increased in the ratio of their use. Undertaken by the writer for the 
purpose of standardizing work in our own office, the task has developed 
to a point where it seems apparent that the results will be useful to 
many designers, and are, therefore, made available by publication. 

There has also been given a brief summary of the best comparative 
information available to the writer with regard to methods in general 
use by designers, and the results of experimental determination of 
stresses by test, for the purpose of giving anyone using the diagrams 
a correct sense of the meaning of the results obtained with reference 
to other methods. Also comparative information is given as to the 
factors of safety, which exist in connection with the use of working 
stresses, as defined by Ordinances and other authorities, in contrast 
with ultimate values obtained by experiment. 

The need of such general information is amply shown in the data 
given with regard to bending moments. 

Primary assumptions used quite generally by practical designers 
are shown to vary as much as one hundred per cent, and give results 
at variance with each other to the extent of fifty per cent. 

The range of opinion and practice with reference to the calculation 
of reinforcement, for diagonal tension is in even a more chaotic state. 

The principal diagrams are presented as two-line nomographs, 
which, in the form used, furnish a solution of what would be a series 
of lines on the ordinary graph, by the use of only two lines, thus 
reducing the objectionable features, caused by the complexity of 
numerous lines on the ordinary graph, to a minimum. 

These diagrams are to be considered as a special form of slide-rule 
devised for the particular purpose in question, which permit the 
solution of problems by the use of standard formulae in any way 
the user may see fit, only governed by the requirements of local 
Ordinances. 

Most of the formulae are in the form used by Professor Arthur N. 
Talbot of the University of Illinois. This is, of course, due to the 
circumstance that the only tests on column footings that have ever 
been made are those described in Bulletin 67 of the University of 



Z HICKS REINFORCED CONCRETE FOOTINGS 

Illinois, carried out under his direction. It, therefore, follows that 
the notation used by him will be best adapted to the purpose in hand, 
and where in a few cases it is departed from, both notations are given. 

Where other formulae are required, general in character, those 
recommended by the Joint Committee* have been used. 

The diagrams are entirely general in nature, and care has been used 
to prevent the intrusion of anything not warranted by competent 
authority. They may consequently be considered as tools, which will 
enable their user to avoid the laborious arithmetical computation 
connected with the use of footing formulae, in exactly the same 
manner that the slide-rule makes it possible for one to use mathemat- 
ical expressions with greater facility and saving of time than is 
possible without it. While a slide-rule may be considered an adjust- 
able nomograph, the specialized nomograph has a further advantage 
not possessed by the ordinary slide-rule, of eliminating to a very great 
extent arithmetical error in the results of any given problem. Since 
the multiplication of constants has been effected and recorded in the 
form of a smooth curve, the chance of error in any problem is 
minimized in direct proportion to the number of operations eliminated, 
in the performance of the work. 

It will be seen that it in no way supersedes an understanding of 
the formulae themselves any more than would be the case with the 
slide-rule, but it does tremendously enhance one's understanding of 
the subject by increasing the rapidity with which he gains an intelli- 
gent conception of the effect upon his results of any given variation 
in the factors. For this reason it will be found useful, not only to the 
expert, who does not need the simple facts presented for the benefit of 
the less sophisticated designer (but whose facility of action will be 
increased by the use of a better tool), but it also furnishes the 
beginner with reliable methods which make it possible for architects 
and engineers entrusting design to subordinates to save much wasted 
time in securing results that are correct and satisfactory. It also 
gives to them the power to check results here and there without loss 
of their own time. 

It likewise affords valuable assistance for rapid analysis to the 
examining authorities of Building Departments, and to students and 
teachers, for purposes of instruction or the working up of experi- 
mental data. 



*The latest report of the Joint Committee on Reinforced Concrete appears in 
the Proceedings of the American Societj' of Civil Engineers for February, 1913. 

The Committee carries representation and authority from the American Society 
of Civil Engineers, the American Society for Testing Materials, American Railway 
Engineers and Maintenance of Way Association, and the Association of American 
Portland Cement Manufacturers. Its recommendations are the careful findings of 
experts best qualified in their several lines to establish limits, and its report 
unquestionably represents the best authority extant on the subject. 

For the sake of uniformity of practice, everyone desiring to promote rational 
design will support its conclusions until they are changed by further experience 
or replaced by better authority. 



INTRODUCTION 3 

The nomographs have been printed en cloth, on a larger scale, 
separate from the book itself, so as to be available on the draughting 
board, and after having acquainted oneself with the contents of the 
booklet, it will only be needed for reference, and need not incumber 
one's working space. 

The writer desires to use this opportunity to express his sense of 
obligation to Professor Talbot for the thoroughness with which the 
data contained in Bulletin 67 has been presented, and in this connec- 
tion to say that the results herein given are original only in their 
presentation of methods for using the information made available to 
designers by these important experiments. Whatever value attaches 
to them is merely that they furnish a practical interpretation of the 
work of Professor Talbot. 

That the matter herein presented may prove useful to designers, 
may insure more rational footings, thus tending to unify practice, and 
may render effective assistance in promoting the use of information 
provided to the profession by Professor Talbot in Bulletin 67, is the 
purpose of its publication. 

L. A. H. 

San Francisco. Calif. 
May, 1915. 



4 HICKS REINFORCED CONCRETE FOOTINGS 

NOTATION. 

Square Concrete Footings. 
A used as a superscript indicates that value of symbol affected is to 



be found in the diagrams. 
diagrams. 



TALBOT. 



For dimension symbols see draiuing of typical 
footing on page 7. 

w Allowable soil reaction, pounds per sq. ft. 

W Load to soil from column and approximate weight 
of footing. 

E One side of square concrete column, shoe, C. I. base 
or steel slab. 

L One side of square footing. 

E/L Ratio between widths of footing and concrete col- 
umn or other load transmitting element. 

B One side of square cap. 

B/L Ratio between widths of cap and footing. 

Projection of footing beyond cap. 

Combined depth of footing and cap above steel. 

j Distance from center of steel to center of com- 
pression. 

d Depth of footing from upper face to center of steel. 
d/L Ratio of depth to width of footing. 



LB 

2 
H 



M 



m 

M s 
M c 
A a 


As x 
o 

Av 8 

N 



Bending moment, or equivalent width of beam in 
which steel for bending moment is distributed. 

No. of bars used for bending moment, and to calcu- 
late Bond Stress. 

Resisting moment of steel in width M. 

Resisting moment of concrete in width M. 

Area of steel in width M — Sq. ins. 

Width beyond M on either side at corners of footing. 

Area of steel used in width 0. 

Perimeter of one reinforcing bar. 

Area of steel used for stirrups, as reinforcement against diagonal 
tension. 

Width within which any bent bars used for reinforcement against 
diagonal tension are located. 



NOTATION 



Av b 



Vn 



V n 



DIAGRAMS. 

Area of steel used for bent bars in width N. 

External Shear in width E for depth H, or in width B for depth 

d on one side of footing only. 
Allowable unit resistance to shear at same points. 
External shear at any point 8 t selected as place of measurement 

for diagonal tension, other than E, B, or S. 
Unit resistance to shear, caused by V n . 



S 



Ur iir vertical sheor or emu poinr between 
fo:e of cap and 5 . used as a measure 
of] diagon al tension. 




S. Fbint" of ao lb shear 

5, point at which vertical Shear is w&ed as 

on tnde* of diaoonal tension . 
2) Distance from 5 to S|j 



Fig. 1. Illustrating Symbols 

Used in Diagonal Tension 

Formulae. 



S/L Ratio between length S (= one side of square where unit shear 

equals 40 lbs.) and size of footing L. 
Si/L Ratio between length 8 t {= one side of square at point selected 

as place of measurement for diagonal tension) and size of 

footing L. 
y/L Ratio between distance y (distance between point selected at 

which to measure shear as an index of diagonal tension and 

point of allowable 40 lb. shear in concrete) and size of footings 



S/L - Si/L 

I L 



u a Average unit bond stress. 

f g Allowable unit stress in steel. 

f c Allowable unit stress in concrete. 

k Distance from Neutral Axis to top of footing. 

Other letters as shown on drawing of typical footing, page 7. 



TIICXS — REINFORCED CONCRETE FOOTINGS 



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b hicks — reinforced concrete footings 

Designation of Lines on Diagrams. 

All diagram problems are solved by following line marking desired 
point of use, on scale of origin, horizontally or vertically as case is, to 
its intersection with the curve relating to the problem; thence at right 
angles to first direction to scale relating to answer. 

Arrows at point, where designating letters are placed on curves, 
indicate direction of movement from scale of origin to scale for answer. 

Scales are referred to as "Right Scale," meaning scale on right 
hand of chart ; "Left Scale for B/L" — E/L or d/L, all using the same 
scale; "Upper Scale for w" and "Lower Scale for L," etc. 

Depth. 

K Factor for variation in soil pressure w. 
C Factor for variation of cap width B/L. 

Area of Steel. 

P Factor for variation of cap width B/L. 
A Factor for variation of footing width L. 

Diagonal Tension. 

S Point of 40 lb. unit shear with reference to 120 lbs. at face 

of cap, for variation of B/L. 
Sj Point of 60 lbs. unit shear with reference to 120 lbs. at face 

of cap, for variation of B/L, or any point between 8 and 

face of cap. 

B Face of cap for any B/L. 

Intercepts between lines 8 and S 1 on line of B/L used 
give value of y for one value of v n . For any value of v „ 
Intercept y may be obtained from diagram in manner 
explained elsewhere. Intercept on line of B/L used, 
I: between lines B and S, give distance of point of 40 lb. 

shear from face of cap for use' in fixing position of web 
reinforcement. 
Bond. 

Curves of equal bond stress = 100 lbs. designated by the size of 
the bar, have been placed inconspicuously on the chart 
so that size of square deformed bars is determined at 
a glance. 

Also methods of using the diagram curves in connection with 
table of bar data to adapt results to deformed round bars, 
plain square or round bars, and to establish the amount 
of tension steel that can be bent up within maximum 
bond limits, are given in explanations. (See Page 21 ) 

Limit Stress in Concrete. 

Z Where values of soil pressure iv and cap width B/L inter- 



DIAGRAM EXPLANATIONS 9 

sect above this line, footing depth will be governed by- 
limit stress of 650 lbs. in concrete. For intersections of 
these factors below this line, the concrete stress will be 
less than 650 lbs. and need not be calculated. 

External Shear. 

V While not required for standardized design, examination 
and analysis may sometimes make it convenient to know 
the External Shear at any point in a footing. This is 
given by the expression 

V- V A W 

Bending Moment. 

M While not required for direct calculations, since it is inte- 
grated into the curves for determination of Steel Area, 
it becomes necessary when depth is determined by stress 
in concrete and is convenient for purposes of analysis. 
The desired value is expressed 



Working Stresses and Diagram Constants. 

Allowable Unit Working Stresses. 

Tension in steel 16000 per sq. inch 

Shear in steel 11200 " " " 

Max. Fiber Stress in Concrete 650 " " " 

Max. Vertical Shear in plain concrete .... 40 " " " 

Max. Vertical Shear in reinforced concrete with 
diagonal tension provided for by reinforce- 
ment 120 " " " 

Bond Stress — 

Plain Bars 80 " " " 

Deformed Bars 100 " " " 

Bearing on Concrete 500 " " " 

Value of Constants. 

j, .875. k, .375. 

Where it is desirable for purpose of analysis or experiment to use 
more accurate values of these constants, the diagram results containing 
these factors may be multiplied by the constant used and properly 
affected by the more correct value. 



10 hicks reinforced concrete footings 

Formulae. 



WORKING TRANSFORMATION. 



REFERENCE. 



, . . _ -. / W SIZE OF FOOTING. 

(1) L = V l 



w 

PUNCHING SHEAR. 



(2)d-C A K A L a. (L*- 3^w 



(4Bj)(l20) 4Bjv 

AREA OF ONE WAY TENSION STEEL FOR WIDTH M. 

(3) A s = P* A * A ^ (jac 2 4-.6 c a )w 

S *(l6000)(7/8)(d) 



M.-R-I-SM4. L - (B + 2d) 
2 



AREA OF ONE WAY TENSION STEEL FOR WIDTH O. 

(4) A B1 - (*) (°) A( 



Fractional coefficient % may be replaced 
with any value less than unity, dictated by 
the judgment of the designer. 

VERTICAL SHEAR AS A MEASURE OF DIAGONAL TENSION AT ANY 
i POINT /Si GOVERNED BY ORDINANCES OR AUTHORITY. 

(5) ^ V* W EXTERNAL SHEAR. y== ( ac + C 2 ) W 



(6) Y n~ ^ ^ . UNIT SHEAR. y _, V 



Si jd 



(7) 



AREA OF STEEL FOR DIAGONAL TENSION. 
General. 

144 (ZB^_4Q) ^^1 ^ (y) (j*) 



Av«= ^— 2 1 A Va = X-8 



^8 



16000 



Av8= 771* 



FORMULAE 



11 



If Vn — 40 is Jess than 40 lbs., no diagonal tension, beyond the allowable stress 
taken by the concrete, exists at the point selected. If some minimum amount d* 
required and used, as for instance at the point S L , where v a = 60, formula may 
be written — 

Minimum. 

144( 60 - «0 ) (8+gl \ (.05) (L 2 ) 



A Vs = 



2 2 

T3OT5 



or 



(8) 



Av 8 = .0023 (-Ml) L* or(|+|l) L a 



444 



For Y a 120 the formula may be expressed 

Maximum. 

144 ( 130 - gO V| +gl\ (.15) (j 
2. 



(9) 



16000 
A V8 = .027(| + |1) (1 «) 



or 



(10) Av h = 0.7 ir, Bent bars at 45°. 

AVERAGE BOND STRESS. FOR ONE BAR. TENSION VALUES 

@ 16,000 LBS. PER SQUARE INCH. 
Square Bars. 

(11) Tension - (1000) (0*)= lou 

Round Bars. 

(12) Tension = (1273) (0*)= lou 

EMBEDMENT FOR 45° BENT BARS. FROM LINE OF DIAGONAL 
TENSION RUPTURE TO UPPER CLEARANCE BEND. 



g • /IvailaDie bond lenqth above line of ruprorc. .ol 

c ^ sreel clearance ^j 

x - Polnr where decrease. of bend- ^ 
inq momenr permirS bendinq bars.v^ 

g. \Ai{d-c) -g.7(d-x) 




Fig. 2. Illustrating Formula 
13. GRir for Bent Bars. 



Let available grip length . . . . =g. 

Clearance — c. 

Distance cf bend from face of cap = x. 



12 HICKS — REINFORCED CONCRETE FOOTINGS 

(13) Then ^ = 1.42 (d — c) — .7 (d — a;). 



(14) 



(15) 



MAXIMUM BOND STRESS. 



mojd 



CONCRETE STRESS TO DETERMINE DEPTH. 



d= 



_ 2M 



650 kjb 



Use of Diagrams for Design. 

A general problem is assumed, and the explanations for the use of 
curves is followed by specific application to the particular problem. 

The problem to be solved is as follows, and reference is made to the 
typical footing on Plate 2, page 7, for symbols used : 

Design a square footing for a plate and angle column resting on a 
cast iron base, 3' 6" square. The load delivered to the soil by the dead 
and live load carried by the column inclusive of the approximate 
w r eight of the footing itself is 700,000 lbs. 

Allowable soil pressure has been determined as 7,000 lbs. per 
square foot. 



Width. .jfcen L= n/ 700 ,000 _ 10 , „ 

V 7,000 

Depths. To ascertain the depth required for 120 lbs. shear through 
the cap and footing at the edge of C. I. base, find ratio between width 
of C. I. base and one side of footing, or 

E/L- ^-,= .35 
10 

Opposite .35 on Left Scale, find point on Curve C, ver- 
tically below 1.48, the answer, on Upper Scale for C. 

Below 7000 on Upper Scale of soil pressures w, find 
point on Curve K opposite .199, the answer, on Left Scale. 

Then H = C A K A L= (1.48) (.199) (10' 0") = 2.95 ft. 

Results may be recorded in typical footing schedule, as on page 7. 

To ascertain the depth required for 120 lbs. shear through the 
footing alone around edge of cap, find ratio between width of cap and 
footing, or 



USE OF DIAGRAMS — DESIGN 13 

BA- 4iiU. 45 

10 

Opposite .45 on Left Scale, find point on Curve C ver- 
tically below 1.04, the answer, on Upper Scale for C. 

The value of K for 7000 lbs. soil pressure has already 
been determined, as .199. 

Then d=C A K A L= (1.04) (.199) (10' 0") -=2.07 ft. 

D = d + 3 == 2.32 ft. 

I = H — d = 2.95 — 2.07 = 0.88 ft. 
Area of Steel. 

Opposite B/L for width of cap used, viz., .45, on Left 
Scale, find point on Curve P vertically above .999, the 
answer on Lower Scale for P. 

Above width of footing 10, on Lower Scale L, find point 
on Curve A opposite value 7.15, the answer on Right 
Scale. 

Then A s = P A A A = (.999) (7.15) = 7.13 sq. ins. 

Size of bars for tension steel will be decided upon when bond con- 
ditions are considered. 

Width of Footing in Which A s is to be Uniformly Distributed. 

M= B+ 2d + L -B+2d 
2 

4.5 + 4.14+JL0_=_8i_S4_« 9.32 Ft. 
2 

_M ^ 10 - 9.32 c 0.34 Ft. 



2 

Area of Steel for Width O. 

A 81 = (*) (£) A 8 

-(4) Q -34 (7.13)= 0.18 eq.ine. 
~~^* v 10 

The factor 3 4 may be changed by the designer to any value less 
than unity, which in his judgment is warranted. 

Diagonal Tension. The point selected at which to measure vertical 
shear is Id from face of cap = 2.07 ft. 



14 HICKS REINFORCED CONCRETE FOOTINGS 

Opposite B/L used for cap width viz. .45, on Left Scale, 
find point on Curve S vertically above value .745, the 
answer on Lower Scale S. 

Then S/L = .7-45 or flf = 7.45. 

S represents one side of square, where the resistance to shear equals 
40 lbs. The length of one side of square at.point selected for diagonal 
tension index will be 

8 j = 4.50 + 2.07 + 2.07 = 8.64 ft. 

As S t is greater than S, the vertical shear at point selected will be 
less than 40 lbs. and need not be calculated. 

As Joint Committee report of 1913 allows 120 lbs. vertical shear 
"when tension normal to shearing plane is provided for by steel," 
some minimum amount should be used, and it is proposed to make the 
calculations for 20 lbs. of shear in excess of the 40 lbs. allowed in the 
concrete to illustrate the use of diagrams, leaving to the designer's 
judgment and the requirements of local codes the decision as to what 
is allowable in practice. 

Using Formula (7), 

S/L as found above = .745. 

Opposite B/L used for cap width, viz. .45, on Left Scale, 
find point on Curve Si vertically above value .65, the 
answer on Lower Scale S. 

On the line of B/L = .45, the intercept distance between 
Curves Si and S may be counted directly in unit sub- 
divisions, and is found to be 4.9 units, or y = .049, or may 
be figured as — 

SA - Si/L _ .745 - .65 - .0475 
2 "3 " 

Then using Formula 7, 

144 ( 60 - 40 )^ .65 +.745 ) ( . 0475) (l0 2 ) 

H v B ^ 16000 « 

(144) (10) (.6975) (.0475) (10) (10) . 0.3 flq.ins. 
16UUU 

or about 5% of area of tension steel for M. 

If bent bars are used, the area required will then be 

Av b = (.07) (Av,) = (0.7) (0.3) = 0.21 sq. ins. 
or about 3% of steel required for tension in width M. 



USE OF DIAGRAMS DESIGN 15 

Where the point selected for an index to diagonal tension is such 
that the total shear is 60 lbs., as in this case, the shorter formula (8) 
may be used, viz. : 

A Vfl « .0025 (S+ Si) L 2 

^(.0023) (.65-K745) (100)= .32 eq.ins. 

Assume that the design is governed by the condition, that vertical 
shear % d from face of cap is to be used for diagonal tension. 

TVi^n s l - 4*50+ 2.07 _ .657 
men — j- jxj - 

As this is practically the same value as that already used, the 
calculation will not be repeated, but the work should be done with 
the general formula ( 7 ) . 

If reinforcement is required for the entire vertical shear in excess 
of 40 lbs., use formula (9). 
Then 

Av s = .027 (.745 + .45) (10) (10) = 3.22 sq. ins. 
Bond Stress. For Square Deformed Bars at 100 lbs. 

Opposite B/L used on Left Scale, find point directly 
above size of footing on lower scale L. Observe the next 
line of equal bond stress above this point. This repre- 
sents the size of bar that will afford an average unit bond 
stress not greater than 100 lbs. In this case %" square 
bars would be indicated. 

Bent Bars. It has been shown that if bent bars are used, the area 
required amounts to .21 square inches, or 3% of the main tension bars. 
As these bars can be bent up with little increase of cost and as experi- 
ment shows much greater ultimate security with effective web rein- 
forcement, it is always desirable to bend up an adequate number of 
bars to take the full estimated diagonal tension. The effect of this on 
the Bond Stress in the remaining bars must be determined. 

Observe that the perimeter of % bars is 3 (See Bar 
Data table), that of a % bar 3.5, and that the intersection 
of B/L and L is approximately one-fifth of the linear dis- 
tance between the %" and % lines, below the % curve. 

Then the perimeter of the bar that would be adapted to give 
precisely 100 lbs. adhesion is 

3+ (1/5) X (3.5 — 3), or 3.1. 



16 HICKS REINFORCED CONCRETE FOOTINGS 

The next line above the % line will probably provide sufficient 
surplus bond to allow of bending up the number desired without 
exceeding 100 lbs. in the remaining bars. The number of bars required 
for bond, of the size proposed for use, viz. %", will be found accurately 
as that part of the number required for tension represented by a 
fraction having as numerator the perimeter of the bar to be used, and 
as denominator the perimeter of an imaginary bar adapted to the 
development of 100 lbs. bond at the B/L used. 

Illustrating : 

% sq. bars required for tension, 7.15/.391 = 18.4. 

Bars required for bond, (2.5/3.1) (18.4) = 14.4. 

Or 4/18.4 = 22% of tension steel may be bent up. 19 bars will be 
used in tension, and 4 bars for diagonal tension. 

The diagram of bending moments shows that for a considerable 
distance from face of cap, the bending moment is decreasing at the rate 
that makes it possible to dispense with this amount of steel 5" from 
face of cap. See page 7 for Bending Moment Diagram. 

The calculation is very simple and is as follows : 

2/3 of projection C = 2/3 of 33 = 22 inches. 

Then the rate at which the bending moment is decreasing per inch 
away from face of cap will be : 

l/22 = 4y 2 % per inch, or 22%% in five inches. 

As one bar represents 1/19% of the area actually. used, or 5*4%, 
the four bent bars will represent 

(4) (5y 4 ) =21% bent up. 

Round Deformed Bars. If round bars are to be used, multiply the 
perimeter represented as the intersection of B/L and L by .7854. This 
gives an imaginary perimeter suited to the development of 100 lbs. of 
bond and to be used as the denominator of the fraction to be similarly 
determined as with square bars. 

For example, (3.1) (.7854) =2.44. 

As 2.36 (See bar Data Table) is the perimeter of % bars, % bars 
will probably provide the surplus bond required. Then 

% round bars required for tension, 7.13/.31 = 23. 

Bars required for bond, (1.96/2.44) (23) =18.5. 

Or 4.5/23 = 19%% of tension steel may be bent up. 

If this is not sufficient, use the next smaller size, and when a con- 
siderable amount of tension steel is to be bent up designate two 
points — for bending proportioned to the decrease in bending moment. 

If small bars become objectionably numerous, resort may be had to 
vertical stirrups in place of bent bars, or to any desired combination of 
the two methods. 



USE OF DIAGRAMS DESIGN" 17 

Above the curve for % square any type of mesh may be used rather 
than plain bars. 

Plain Bars. As pointed out later, deformed bars at working stress 
of 100 lbs. give much better factors of safety than plain bars at 80 lbs., 
and the method of deriving the necessary results where plain bars are 
used is, therefore, given in' connection with its use for analysis in 
Part 7. 

Bond Stress. Bond stress in bent bars may be provided for by 
extending the bar to grip length beyond the 45° line of diagonal 
tension fracture equivalent to its tension value, but the cheaper, easier 
and safer plan is to specify that all bent bars be hooked at the ends 
with standard hooks, which are capable of developing the tension 
value of the rods. 

Stirrups. Talbot concludes that .6 the depth from the surface of 
the concrete may be used as part of length for grip adhesion. This 
length plus any added bends or hooks at the top must supply resistance 
to bond stress equivalent to tensile value of the stirrups. 

Areas, perimeters and tension values are given in table of bar data. 

If stirrups are made continuous, bond stress becomes negligible. 

Design. The use of the diagrams for the design of a typical square 
footing will now be illustrated without repeating same explanations. 

Soil load 5000 lbs. per sq. ft., column load and footing weight 
390,000, carried by spiralled concrete column 22" square. 



Then L= -\/ 390,000 ^ 8-83 ft. 
V 5000 
Make cap 6" wider than column. 
Then 

Ratios E/L = 22"/8.83 X 12 = .208. 

B/L = 34/106 = .32. 

H=CKL= (2.72) (.14) (8.83) =3.36. 

d=^CKL= (1.66) (.14) (8.83) =2.06. 

D = 2.06 + .25 = 2.31. 

I = H — d = 3.36 — 2.06 = 1.30. 

The minimum depth for /, or depth of cap, will always be deter- 
mined by the limit shear, as 120, at the face of column, through the 
combined depth of cap and footing. 

The maximum depth for I will be any thickness in excess of this 
minimum required by code provisions or Building Department inter- 
pretation. 

j 
M= 2.83+ 4.12+ 8 * 83 | 6 * 95 ~ 7 * 89 ft * 



18 HICKS REINFORCED CONCRETE FOOTINGS 

A s = PA = (.90) (5.60) = 5.04 sq. ins. 

Intersection of B/L = .32 and L = 8.83 is 4/10 below the %" 
curve. Perimeter value for 100 lbs. bond is, therefore, 3.4. % bars 
with 2.5 perimeter and area of .39 will be used. Then 

Bars required for tension, 5.04/.39 = 12.9. 

Bars required for bond, (2.5/3.4) (12.9) =9.5. 

Then 3.4 bars, or 27% of steel, may be bent up. 

13 % square deformed bars 8' 6" long will be used each way, 3 of 
which, or 23.6%, will be bent up. 

Width of projection 

r, g 6.83 - 2.83 = s» o M = 36" 
2 

2/3 of = 24". Then from Bending Moment Diagram, bending 
moment diminishes 1/24 = 4.17% per inch. 

Then 23.6/4.17 = 5 2/3, or say 6", the distance from face of cap, 
where 3 bars may be bent up. 

Ordinance allows point at distance jd from face of cap to be used 
as place to measure vertical shear for diagonal tension, or 

(%) (2.06) = 1.81 ft. from face of cap. 

Opposite B/L .32 on Curve S find point above .635 on 
Lower Scale. 

Then point of 40 lbs. shear occurs at 

(.635 x 8.83) - 2.83 „1.39> 

from face of cap, and shear reinforcement will not be required by 
ordinance, but will be used as before figured, as it practically costs 
nothing, and adds much to security. 

Bond in Bent Bars. Using Formula (13) 

Available grip length 

1.42(2.04 — 2")— .7(2.04 — 6") 

1.42(1.88)— .7(1.54)= 2.67 — 1.08 = 1.59. 

19 
1.59 = 19". Then — - = 30 bar diameters. 

% 

As 40 bar diameters are required for bond, an additional length 
of 10" bent down parallel with upper surface of footing may be used, 
or Standard hooks requiring a bend of 180° through a diameter of 3" 
may be specified. The hooks and a short tangent at ends will require 
6" in length of rods. 



USE OF DIAGRAMS DESIGN 19 

The length of the three bars to be bent up may now be estimated 
as follows : 

Width of cap 2.83 

2X=(2) (6")= Dist. to bend from cap . . 1.00 

2 X 2.67 inclined part as above 5.34 

2 hooks @ 6" each 1.00 



Total length 10.17 

Tension bars for width 

0= 8,83 - 7*87 = 43 
2 

(%) (.48/8.83) (5.04)=. 21 sq. ins. 

Use y 2 " square deformed bars area = 0.25. 

Concrete Stress. 

As intersection of B/L .32 and soil pressure 5000 occurs 
below line Z , this will be less than 650 lbs. and need not 
be figured. 

No stirrups required. 
Spacing of tension bars: 

One bar at either end of M leaves 12 intermediate spaces, or 
7.87/12 =.6o6, or slightly less than 8". 

Bars in width O will be centered or placed half way between last 
bar in M and edge of footing, leaving about 3" from these % bars to 
face of concrete. 

For further information as to the application of diagrams to con- 
ditions not explained, consult the last section of this booklet. 



20 



HICKS — REINFORCED CONCRETE FOOTINGS 



Use of Diagrams for Review. 

Building Departments. 

The adaptation of these diagrams to review assumes the ready 
determination of stresses existing in designs presented for approval. 

The examining authority is only interested in assuring itself that 
the code limits are respected, and is not at all concerned as to how 
wasteful the design may be within those limits. 

The methods presented, therefore, are directed to the determination 
of correct dimensions for the loads assumed, and comparison of these 
with actual dimensions of the design. 

Examples of actual footings selected at random from designs taken 
from Building Department records of Western cities will serve to illus- 
trate the facility with which the desired information may be secured 
by the aid of diagrams. 

An official report blank covering the points to be investigated will 
insure systematic record of results, and avoid the omission of anything 
requiring consideration. 

Such a blank adapted to local needs with regard to working stresses 
used will contain place for sketch and heading as shown, and for any 
other desired data. 

No. 1. Foundation Report Building. 

Date Street 

Owner Architect 

SKETCHES. 




Fig. 3. Review of Col. E 5 
Building. 



USE OF DIAGRAMS REVIEW 

Footing No. E 5. 
O.K. 
Soil pressure 6350 lbs. per sq. ft. Value of # = .18. 
Size. 



■V 



334000 



6350 



7» 3 W 0. K. 



Depths. E/L = (20/7' 3" X 12) = .23.B/L=(2.67/7.25)=.368. 

Cap and Footing. H = CKL= (2.44) (.18) (7.25) = 3.18 

Design 2.63 Inc. 
d =CKL= (1.39) (.18) (7.25) = 1.83 

Design 1.46 Inc. 
Shear at Face of Column. (3.18/2.63) (120) = 145 Decrease. 
Shear at Face of Cap. (1.83/1.46) (120) == 149 Decrease. 

Area of Steel. A s = PA = (.955) (3.8) = 3.63 sq. ins. 

Design, 9 plain 1%" each way, (9) (1.23) ^ 11.04 O.K. 
Width M. 

(2.67)+(2) (1.83)+ 7 * 25 * 6 -3! = 6.79 ft. 

Bond. From Diagram, intersection of B/L = .368 and L = 7.25 
occurs on bond line for % square bars with 2%" perimeter for 100 lbs. 

Correct perimeter for round bars at 80 lbs. 

(2.5) (80/100) (.7854) =1.57, or i/ 2 " round bars. 
See Bar Data Table. 

Perimeter of bar used in design, 3.94 for l 1 /^" 0. Area 1.23. 

Then bars required for tension 3.63/1.23 = 2.95 bars. 

Bars required for bond, (3.94/1.57) (2.93) =7.40 bars. 

As 7.4 bars are required in width M, and at least 1 bar must be 
placed on each side in width O, it appears that 9.4 bars are needed, 
and that no surplus bond permitting bending of part of bars exists. 

As no significant expense is caused by requiring bent bars, the 
design should be rejected. 

This was at once apparent from the total area of steel and size of 
bars used. The calculations have only been continued to show the use 
of diagrams in analysis of an uneconomical design. 

The data will now be used to illustrate diagram bond design, and 
also economy arising through the selection of type of bar. 

Since the correct perimeter for plain bars at 80 lbs. is shown to 
be 1.57, and bent bars are required, the next size will be used, viz., 
% having an area of .11 sq. ins. and a perimeter of 1.18 ins. 

Then bars required for tension, 3.63/.11 = 33. 



22 HICKS — REINFORCED CONCRETE FOOTINGS 

Bars required for bond, (1.18/1.57) (33) =24.8. 

8.2/33%, or 25% of steel in width M, may be bent up. 

Then using 33 bars for 32 spaces, the centering will be 2.55 inches. 

While the allowable interval between bars (as fixed by the Joint 
Committee) is not exceeded in this case and will rarely be reached in 
footings, the large number of bars increase expense of placement, and 
involve higher tonnage price. 

This leads to the general conclusion that square deformed bars 
represent maximum economy, under present conditions of allowable 
working stresses, and the actual relations of areas and perimeters. 
See page 57 for verification. 

The results of substituting square bars mark the contrast. 

Correct perimeter 2.5. 

Use next size perimeter 2. Area .25. 

Bars required for tension, 3.63/.25 = 14.5. 

Bars required for bond, (2/2.5) (14.5) =11.6. 

This requires the handling of 15 bars each way in place of 32, with 
a direct saving of about 50% of placement labor, a slight saving in 
tonnage cost due to using a y 2 " square instead of a % 0, and a possible 
small increase in using deformed instead of plain sections. In the 
San Francisco market no differential is now made between plain, 
twisted and deformed bars. 

Also a more satisfactory centering is secured, viz., 15 bars for width 
M with 14 spaces @ 5.8" centers. 

Diagonal Tension. If diagonal tension were now investigated on 
the basis of the fact that the city ordinances permit vertical shear at 
distance of y 2 d from face of cap to be used as index of diagonal 
tension, the following results would be obtained : 

From Diagram : Opposite B/L = .368, find point on curve S 

S 
vertically above answer .68 on Lower Scale S, being value of — - 

Value of 

Si=,B+-|* = 2.67+1.82=4.49 

Then _Sl = 4 « 49 = .62 
L 7,25 

For External Shear. Opposite .62 on Left Scale, find point on 
curve V vertically above .153, the answer on Lower Scale V. 
Then V=V A W= (.153) (334000) =51200 lbs. 

Vst 512QQ -s go lbs. 

(4.49) (12) (7/8) (1.83) 



USE OF DIAGRAMS REVIEW 23 

As 40 lbs. is allowable in concrete, design will call for reinforce- 
ment for 20 lbs. in steel and Formula (8) may be used. 

A Yo - (.68+ .62)(7.25 z ) _ .16 eq.ins., or 

444 

.16/3.63 = 4.4% of tension reinforcement. 

Previous calculations show that 2.9/14.5, or 20%, of steel may be 
bent up, or practically four times the amount required, for vertical 
steel. 

Bending moment diagram shows that 20% of steel may be bent up 
4 inches from face of cap. 

Calculation for this information follows : 

C= 7 - 25 ~ 2 - 67 =2 .29 = 27i/ 2 ". 

Li 

2/3 of C = 18.33 inches. 

Inspection of bending moment diagram shows that bending moment 
lessens at such a rate that it will have decreased 1/18.33 = 5.4% per 
inch away from face of cap. The distance away from projection for 
207c of steel to be bent up will then be 20/5.4=3.7 inches, say 4 inches, 

Talbot's experiments show that this steel should be used to guard 
against failure by diagonal tension, which may occur in reinforced 
footings, when the external loads are such as to produce elastic limit 
stresses in the tension steel. 

At working stresses there is no stress whatever in the diagonal 
tension steel, for the concrete carries these stresses until incipient 
cracks begin to transfer them to the steel. 

It follows, therefore, that the most advantageous location for such 
steel is normal to the plane of fracture and midway in its length, or on 
the plane of symmetry passing through the center of the footing 
section. 

This would imply bending at the vertical line passing through the 
face of the cap, but as this is impossible, and as the anchorage does not 
extend to the top of the footing, the point of symmetry is really below 
the center several inches. 

It will be seen from Figure (2) that the condition as to anchorage 
is improved as the point of bending is removed from the edge of cap, 
and as long as the steel crosses the line of fracture fairly near the 
center of the effective depth, its placement may be considered 
satisfactory. 

In actual practice, the investigation need not have continued 
beyond the discovery that the depth of footing is deficient, for this 
relation is fundamental since it enters into all subsequent calculations. 

It follows, therefore, that footings may be of any depth involving 



24 HICKS — REINFORCED CONCRETE FOOTINGS 

less than 120 lbs. of shear, but that no footing can be thinner than is 
requisite for the allowable limit stress of 120 lbs. 

The same data will now be used in modified form to indicate the 
effect on review work, in case the vertical shear is less than 120 lbs. 

No. 2. For same footing, assume that design shows : 

# = 3.25. 
d =2.25. 

A s = 7 %" sq. deformed straight bars, width M. 

3 y% sq. bent bars, width N, bent up at 45° 6" from cap, 

without hooks. 
1 y 2 " sq. bar, width 0. 

Examination. 

Soil pressure 6350 0. K. for locality. 

Size of footing 7' 3" 0. K. for load. 

H = CKL= (2.44) (.18) (7.25) = 3.18 

Design 3.25 0. K. 
d =CKL= (1.39) (.18) (7.25) =1.82 

Design 2.25 0. K. 
Shear less than 120. 0. K. 
Area of steel (in proportion to depths) — 

A s = (1.82/2.25) (.955) (3.8) =2.94 sq. ins. 
Available 10 bars @ .25 2.50 Increase. 

The steel used is about 15% less than the Talbot assumptions, and 
evidently involves bending moment assumptions greater than those 
shown in the comparative diagrams given in Part 5. 

If other than the Talbot assumptions are allowed, the diagram 
results should be affected by a multiplier representing the divergence 
of the assumptions. This may be used as a constant average percent- 
age for all moments, if figured for B/L = .42. 

In this case the diagram results are used, and 2 more bars will 
be required. 

Bond. Diagram intersection % bars, 2.5 perimeter. 
Use y<£ bars, 2 ins. 
Bars for tension, 2.94/.25 = 11.8. 
Bars for bond, (2/2.5) (11.8) = 9.45. 
Used in design, 10. 

Straight, 7. Increase 2. 
Bent, 3. 0. K. 

Distribution. Respace 12 bars in width M = 6.78 ft. 0. K. 



USE OF DIAGRAMS REVIEW 25 

Steel for Width 0. 7.25 — 6.7 8 = .23 ft. 

2 

Nothing required. 

Available 1 i/ 2 " bar. 0. K. 
Diagonal Tension. y>>d from cap. 
Si = 2.67 + 2.25 = 4.92. 

8 = (S/L) (L) = (.68) (7.25) = 4.92. 

As the point of 40 lb. shear coincides with the ordinance position 
for measuring diagonal tension, bent bars would not be required, but 
the Department has ruled that the point of 60 lb. of vertical shear 
shall be regarded as the allowable minimum for figuring diagonal ten- 
sion with web reinforcement provided for 20 lbs. of this amount. 

Then using Formula (8) : 

From diagram for B/L = .'368. 

S/L = .68. S 1 /L = .5S. 

A Yn ^ (.«58+«68)(7.25) Z = .15 sq.ins. 

Amount to be bent up in revised design : 3 y 2 " bars @ .25 = .75 
sq. ins., or 5 times amount required. 0. K. 

Point of Bending. Percentage of steel to be bent up, 3/12 = 25%. 

Bending moment diagram shows that this may be bent up 5 inches 
from cap. 

2/3 C = 18.33 ins. Then 1/18.33 = 5.45 %. 

25/5.45 = 4.6", say 5", from face of cap. 
Design, 6". Reduce to 5". 

Bars shown terminating at point 2" below top of footing. Clear- 
ance 0. K. 

Acting Length of Embedment. 

(1.42) (2.25 — .16) — 0.7(2.25 — .29) = 1.60 = 19%". 
19.25/0.5 bar diameter == 38%. 

Required (Table of Bar Data), 40. 

Bars must be extended horizontally or provided with standard 
hooks. 

It will be seen that most of work involved thus far has been due 
to explanation rather than review. 

Another footing from the same city will now be used : 



26 



HICKS — REINFORCED CONCRETE FOOTINGS 
SKETCH. 



JXk. 



j3q: 



/ 



\ 



\ 



\ 



\ 



/ 



/ 



/ 



/ 



3'4~" 



Zi *le? <f> Z way Every a/hrrnare bar benrop. 



Widrh K» • 8.74-' 











3 


k 




1 








j* 



No. 3 



Fjg. 4. Review of Col. 3 Buildinu. 

9' 4". 0. K. 



- i 



461^000 



5300 

E/L = 24/112 = .214 B/L = 30/112 == .2G8 
H= (2.65) (.15) (9.33) =3.70 

Design 3.58 Increase. 
d =(2.06) (.15) (9.33) =2.85 

Design 2.63 Increase. 
Shear at cap = (2.85/2.63) (120) = 130 lbs. Decrease. 
A s = (.82) (6.25) = 5.12 

Design 21 % @ .31 = 6.52 0. K. 
Spaced 5%" $ each way. 
Every alternate bar bent up. 



USE OF DIAGRAMS REVIEW 27 

M = 2.50 + 0.66 + .oS == 8.74. 

Available in width .¥, 8.T4/.5 = 19. 

As 10 bars are bent up at point of maximum bending moment, only 
9 bars are available for tension = 2.79 sq. ins. Required, 5.12 sq. ins. 
Increase. 

Deficiency in tension steel can be corrected by proper placement of 
available material. 

To ascertain correct distribution, investigate bond conditions. 

From Diagram, bond intersection for B/L = .268 and L = 9' 4" 
occurs at — 

y 8 -\- 2/3 = 3.50 -f- .33 = 3.83 perimeter for square deformed bars 
@ 100 lbs. 

For round deformed bars @ 100 lbs. the correct perimeter would 
be (3.83) (.7854) =3.01. 

Design, % deformed bars with perimeter of 1.96. 

Bars required for tension, 5.12/.31 = 16.5. 

Bars for bond, (1.96/3.01) (16.5) —10.7. 

Reduce number of bent bars and change point of bending. 

As there are 19 bars available for tension in M, and only 11 required 
for bond, 8 alternate bars near middle of footing may be bent up 9 
inches from cap. ft . , rtl A 

2 

2/3 C = 27 1/3 inches. 

1/27.33 = 3.66% per inch decrease of tension. 
Steel decreases 5.5/16.5 = 33.4%. 
Then 33.4/3.66 = 9 inches. 

That is, there must be 11 bars available for bond at a point 9" from 
face of cap. 

Diagonal tension for y 2 d from cap, as per ordinance. 
S x = 2.50 + 2.83 = 5.33. O. K. 

S~(jL) (L)- (.58) (9.33)= 5.41 

Xj 

As St practically coincides with point of 40 lbs. shear, only 
minimum reinforcement for 20 lbs. in steel in excess of 40 lbs. allowed 
in concrete will be required. 

This need not be figured, as the bent bars will provide several times 
more resistance to diagonal tension than will be required. 

Grip Length — Bent Bars. Formula (13). 

1.42(d — c) — 0.7 (d— X). 

1.42(2.83 — .16) —0.7(2.83 — .42). 

3.79 — 1.69 = 2.10 = 2514". 
Diameters of bars — 

25.25/.62S = 40i/ 2 diameters. O. K. 
Critical Points. Examination should be made as to the following 
critical points: 



28 



HICKS — REINFORCED CONCRETE FOOTINGS 



Limit shear at edge of concrete columns, C. I. base, or steel slab. 

Limit shear at edge of cap. 

Bond in tension bars beyond point of bending. 

Bond in bent bars above line of rupture in diagonal tension. 

Bending moment at point where bars are bent up. 

Experiment. 

For the purpose of illustrating the use of the diagrams for the 
calculation of ultimate stresses from experimental data, some of the 
footings of the Talbot tests will be used. 

Since the diagrams are based on constant values of j = .875, 
fs = 16000, v == 120 max., and u = 100 for square deformed bars, it 
is evident that other values of these factors may be obtained by proper 
use of diagram results. 

The first two footings encountered on page 74 of Bulletin 67 will 
be used, as follows: 





L ,0 " 


1505 


-0 


> 










\ 


\ 









1-24-5 Concrete 


1 ' 


b 




3'o" 










1 1506 


b 








.0 













Lines 

Wrfll 



ines* of cro.cK.inq for maximum XZr 
i&tance in homoqeneoys material 
vrHi cap aH-aohed te foohnq Inreqralli 

r~V — I I- cV 5 Concrete 



— A.-1— . 



po.28 15 ^le,"^ Z way 



I4IZ 



Fig. 6. Unreinforced Concrete 
Footings — Cracks at Failure. 
(Talbot.) 



Fk;. 5. Illustrating Cracks on 
Bottom of Reinforced Con- 
crete Footing — Typical Ten- 
sion Failure. (Talbot.) 



USE OF DIAGRAMS EXPERIMENT 29 

Depth 
to 

Foot- Depth Center Cause Size Size 

ing Descrip- Per Load at Over of of of of 

No. tion. Cent. Failure. All. Steel. Failure. Foot. Cap. 

1411 15 % 0.28 112,000 11 10 Tension 5' sq. 1' sq. 

plain 

1412 " 0.28 160,000 11 10 Tension 5' sq. 1' sq. 
Value of j used in Talbot calculations, for p, 0.28 = .90. 

Calculated Values. 
For No. 1411. Soil value 112,000/25 = 4500 lbs. B/L = 1/5 =.2. 
B/L at Id = 2.65/5 =.53. 

For No. 1412. Soil value 160,000/25 = 6400 lbs. B/L = 1/5 =.2. 
B/L at Id = .53. 

For 1411. If cap width were 2.67 there would be vertical shear of 
120 lbs. at its face for a depth 

C K L= (.795) (.127) (5) = .5. 
Actual depth 10" = 0.83. 
Then actual shear Id from face of cap — 

(.5/.83) (120) (.875/.90) =70 lbs. 
Talbot determination 69 lbs. 
For 1412. d = CKL= (.795) (.18) (5) =.71. 

v = (.71/.83) (120) (.875/.90) = 100 lbs. 
Talbot determination 99 lbs. 
It will be seen from the figures that the determination of the depth 
of a footing having a cap of same size as point selected to measure 
vertical shear makes it possible to determine the actual shear as the 
same ratio of 120 lbs. that exists between the depths. Also, since the 
constant ,;' entered into the original calculation as a divisor, it must 
now be put back as a multiplier, and the result divided by the same 
value of j used in the calculations for the experiments. 

It will be interesting to calculate the shear at the cap instead of 
Id away. 

For 1411. C K L for cap ratio .2 = (2.84) (.127) (5) = 1.8 

1412. C K L for cap ratio .2 = (2.84) (.18 ) (5) = 2.55 
Then for 1411. v= (1.8 /.83) (120) (.875/.90) =252 lbs. 
1412. v = (2.55/.83) (120) (.875/.90) = 356 lbs. 
Tension in Steel. Steel required. 

Above 5 on Lower Scale L, find point on curve A opposite 1.8, the 
answer on Right Scale. 

Opposite B/L .2 on Left Scale, find point on curve P above .675, 
the answer on Lower Scale P. 

Then A S = PA = (1.8) (.675) = 1.21 sq. ins. 
Used 15 % @ .11 = 1.65. 



30 HICKS REINFORCED CONCRETE FOOTINGS 

Considered as effective by Talbot : (46/60) (1.65) = 1.265 sq. ins. 
Then 1411. / s — (1.8 /0.83) (16000) (1.21/1.265) (.875/90) = 32200 

Talbot 31900 
1412. / 6 =(2.55/0.83) (16000) (1.21/1.265) (.875/90) = 45800 

Talbot 45600 
Bond Stress. External shear V = V A W. 

Then for 1411, opposite B/L = .20 find point on curve V above .24, 
the answer on Lower Scale V. 

Then v~ *-M ^ 112.000 220 

(15)(.765)(1.18)U0)(.9) 22 ° 

Talbot determination 220 
Similarly for 1412, since all the factors are identical, except the 
loads, the stresses will be as the loads, or 

(160,000/112,000) (220) =314 
Talbot determination 314 
Illustrating the value of average bond stress in contrast with 
maximum stress used by Talbot : 

From diagram, intersection of B/L = .20 and L = 5 occurs 2/5ths 
below the %" line, or (2/5) X (.5) = .2 + 2 = 2.2 perimeter value for 
square deformed bars at 100 lbs. 
For plain rounds at 80 lbs. value : 

(2.2) (80/100) (.7854) =1.38 
Bars used, % 1.18. 
Then for 1411— 

(1.18/1.38) (80) (1.8/0.83) (.875/.90) (1.21/1.265) = 136 lbs. 
Similarly for 1412, since all the factors are identical, except the 
depths, the stress will be as the depths, or 

(2.55/1.8) (136) =194 lbs. 
To check these calculations for the average bond, assume that 
tension steel had clearance at ends of 2", or bars were 60 — 4 = 56" 
long. Then 

(56" — 12) /2 = 22" = effective bond length. 
Tension capacity of one % @ 31850 lbs, 

(1.18) (1.18) (1273) (31850/16000) = 3530 lbs. 
Bond 3530/(1.18 X 22) = 136 lbs. 
Tension capacity of one % @ 45400 lbs. 

(1.18) (1.18) (1273) (45400/16000) =5040 lbs. 
Bond 5040/(1.18 X 22) = 194 lbs. 
The ratios between maximum and average bond are shown by figur- 
ing that the bars extend to the face of the footing, since the external 
shear is used to that point in calculating the maximum stress. 



USE OF DIAGRAMS — EXPERIMENT 31 

Then 5040/(1.18 X 24) =177.5. 

Then Maximum Stress . 314. _ 1.755 
Average Stress 177.5 

This ratio varies frcm 1.5 to 1.77 in the large number of instances 
examined in connection with the diagrams. 

It will, of course, vary with the length of clearance at the end of 
the bars and should apparently be 1.5, when the bars extend to the 
face of footing on account of the parabolic form of the curve defining 
the distribution of bond. 

It will be seen that critical maximum bond stresses are from 1.5 to 
1.77 times as great as the average working stresses used in design. 

When the dimensional constants are worked out for given experi- 
mental information, and the user of the diagrams has become some- 
what familiar with the methods of their application, it will be seen 
that rapid results can be secured. 

The calculations of these and many other experiments have been 
used to check the accuracy of the diagrams, and the results have 
invariably proven within a small margin of error. 



32 hicks — reinforced concrete footings 

General Information. 

The data contained in the following pages is intended to make 
available to beginners in the art the contrasts found in practice 
between fundamental assumptions, the relation between working 
stresses and ultimate rupture loads, as shown by experiment, and to 
interpret to some extent the experimental results used. 

It is substantially the matter prepared for a lantern slide talk 
before the Architectural Club of San Francisco, an association of 
architects and designers, and covers many points of detail in a more 
elementary manner than would be necessary if intended only for an 
audience of engineers. 

Nevertheless, it is believed that many designers will find this form 
of presentation useful, and the information is given in such a manner 
as to enable anyone to make his own interpretation. 

The explanations are, therefore, addressed only to those who find 
them useful. 

Where specific methods are recommended, as in the case of bending 
moment and shear calculations, it should be understood that this is 
only the author's own preference, and that formulae are invariably 
given in such general form as to permit the use of diagrams with any 
assumption desired. 

Caps. The questions relating to the dimensions of caps are usually 
assumed to be so simple as to require no more thought on the part of 
the designer than the use of a 60-degree triangle to establish the depth 
for an assumed projection beyond the edge of column or other loading 
element. 

One of the concrete specialty companies has been largely followed in 
making the projection 6" for columns under 24" in diameter, and 8" 
for columns larger than 24", with depths of twice the projection. 

If the cap is poured separately from the footing, it will be possible 
to determine the stresses in bending and shear. For this purpose a 
24" hooped column will be used with a load of 450,000 lbs. and a 
cap 36x36x12". 

The modulus of rupture for the concrete in the cap will be deter- 
mined by the formula : 
M = sS 
The value of E/L = 24/36 = .666 

Then from diagram for E/L = .666, 

M = (.12) (450,000) (3) (12)/10 = 194,200 
8= (36)(12)(12)/6 = 864 

Then s = 194,200/864 = 225 lbs. per sq. inch. 

Talbot tested a series of concrete footings without reinforcement 
tabulated on page 73 of Bulletin 67. 

Two footings of same depth as the cap just figured are selected 
for illustration. See Fig. 6. 



GENERAL INFORMATION 33 

, 6" Cubes % 

,— Control Beams— >, Max. 

Age Load at Mod. of Mod. of Load per 

No. Days. Failure. Rupture. Rupture. Age. Sq. In. Age. 

1505 86 86,000 195 376 95 3618 128 

1506 77 67,000 152 272 99 2260 116 

Checking the modulus calculated B/L = 1/5 =.2 
For 1505. Ji A from diagram for B/L — .20 == .544 

Then M= (.544) (86,000) (5) (12)/10 = 281,000 

S= (60)(12)(12)/6 = 1440 

5 = 281,000/1440 = 195 lbs. 
For 1506. s will evidently be as the loads, or 

s= (67,000/86,000) (195) =152 lbs. 

Since the conditions of mix and other circumstances were identical, 
there must be a rational explanation as to why No. 1506 failed at a 
smaller load than 1505 and why both failed at smaller unit stresses 
than was the case with control beams. 

It should be remembered, in the case of the footings tested, that 
there is a sudden change in the intensity of the tension stress at the 
face of the cap, because it is integrally a part of the footing. 

Referring to Footing 1506, which failed in a single line of weakness, 
it would seem evident that the failure might have occurred in either 
of four directions as shown in Figure 5. Then if the material were 
absolutely homogeneous it would resist rupture until failure occurred 
simultaneously on all four lines, assuming that the combined resistance 
in these directions were still the critical weakness of the footing. 

If this were roughly so, the resistance developed in 1505 should be 
1.4 times as great as that developed in 1506, or its modulus of rupture 
would be : 

(1.4) (152) =211 lbs. 

The average value assigned the modulus of rupture for concrete 
as determined by control tests is 400 lbs. and effort to rationalize the 
behavior of a material showing such a disproportionate critical weak- 
ness as is exhibited in the low modulus of 152 lbs., at which 1506 
failed, is futile. 

This merely emphasizes the fact that, where concrete is subjected to 
bending stress without reinforcement, safety factors should be taken 
as a proportion of the critical weakness rather than as a proportion 
of the average value of the modulus secured by small tests. 

Returning to the cap considered, two conclusions seem possible : 

First : That in a rough way the attachment of the cap to the 
footing doubles the resistance of the footing, if of homogeneous 
material, since it forces ultimate failure in a perfect material on four 
lines, instead of on two lines in which failure would occur if the cap 
is free from the footing. 



34 HICKS — REINFORCED CONCRETE FOOTINGS 

Second: Caps are subjected to very little bending in actual prac- 
tice, as it is evident that if the footing yielded in the same degree as 
the bed of springs in which footings were tested, the concrete in the 
cap could not resist the bending stresses resulting. 

If these conclusions are accepted, the importance of monolithic con- 
struction between the cap and footing will be appreciated. 

The vertical shear at face of column, considering the caps as sep- 
arated from the footing, will amount to 246 lbs. per square inch, 
as follows : 

From diagram for B/L = 2/3 = .67 

CKL= (.48) (.00002834) (50,000) (3) =2.05 
Then (2.05/1.0) (120) =246 lbs. 

As the value of K in the above example is outside the diagram 
limits, the soil value, 450,000/9 = 50,000, has been multiplied by its 
increment value and so extended to the point used. 

This again illustrates the fact that the cap should be constructed 
monolithic with the footing, and figured for shear at face of column 
by considering the combined depth of cap and footing as affording 
resistance equivalent to external shear within working stress limits. 
This constitutes the basis of design in the diagrams developed. 

It assumes that the compression is distributed through the cap with 
an increasing reduction of unit stress as depth increases, but that the 
result is merely comparable to increasing the size of a column under 
the same loading. Somewhere below the point of such change of 
section, unit stress might again be considered as uniform across the 
entire section, but at intermediate points the outer fibers of the column 
would carry very little stress. 

The cap might also be considered in a way as analogous to a butt 
plate in steel construction. 

In practice the limiting depth of the cap is apt to be some propor- 
tion of the projection allowed by local ordinances or interpretation. 

The rulings on this point in San Francisco agree with the method 
herein used, involving the assumption that the depth of the cap is 
independent of the width of projection and is determined by limit 
shear only. 

This provides minimum depth, which will have to be increased, if 
so required by Building Department interpretations elsewhere. 

In general, the condition of maximum economy will be realized 
when cap width is such that B/L = .60 and cap depth is 4/3 of 
cap projection. 

Depth of Footings. Custom has permitted punching shear around 
the perimeters of footing caps of from 100 to 160 lbs. 

The latest report of the Joint Committee, recognized as the highest 
authority on the subject, allows 120 lbs. for 2000 lbs. concrete "when 
all tension normal to the shearing plane is provided for by rein- 
forcement." 



GENERAL INFORMATION 



35 



If this unit stress is respected, it follows that, with the exception of 
very low soil values and small footings, the depth of all footings will 
be governed by the allowed unit stress in shear. 

As ultimate economy is always in the vicinity of limit stresses, it 
becomes logical to use 120 lbs. as the governing factor in design. 

The failure of footings by direct shear is impossible, as Talbot's 
experiments show failure by diagonal tension at much smaller loads 
than would be required for direct shear. The factor of safety for pure 
shear is probably not less than 10, which assumes a rupture value in 
shear of 1200 lbs. per square inch. 




Fig. 7. Illustrating Diagonal Tension Failure. (Talbot.) 



Nevertheless working limits, when made by competent authority, 
should be respected, and approval of plans should be based on keeping 
stresses within limits assigned by such authority. 



36 HICKS — REINFORCED CONCRETE FOOTINGS 

It follows that intelligent designers should use the full value of 
limits fixed. Employers should not be satisfied to find, in reviewing 
design work, that excess strength has been provided, as is often the 
case where the checker 0. K. 's the results, but should insist that 
proportions be so related that limit stresses in all material will be 
availed of, as closely as possible. 

It will be seen at once that the adoption of a constant maximum 
shear of 120 lbs. means that the ratio of depth to length, or d/L, will 
change with every change of cap size. 

Good designers have used what was considered an average value of 
d/L == .20 as a constant, but consideration of the data in the diagrams 
will show that the range of this factor, far from being constant, varies 
with each change of B/L. 

For instance, for a cap ratio of .48 and a soil pressure of 12,000 lbs. 
the values of C and K are .94 and .343 respectively. Then d/L for 
these conditions equals (.94) (.343) =.322. 

The same data for low soil values and small cap ratios give the 
following results : 

For cap ratio of .35 and soil pressure of 2000 lbs. the figures are : 

d/L=(lA6) X (.057) =.083 

It should, therefore, be remembered that the basis of design in these 
diagrams is the maintenance of 120 lbs. of vertical* shear through the 
footings at the face of all caps. 

On the diagram will be found a line marked Z, above which the 
limiting factor in the determination of depth will be the compressive 
stress in the concrete. 

The intersection of a vertical line from the Soil Value Scale with 
line Z indicates the highest ratio of cap width B/L which can be used 
with that soil value without exceeding 650 lbs. in the concrete, or, 
vice versa, the intersection of any given B/L with line Z is immediately 
under the limit soil value for that B/L. 

It is accurately established for footings of medium width of about 
10 feet, and is substantially correct for all sizes. It will be found that 
depth of very small footings will be determined by some arbitrary 
minimum depth fixed by practical conditions, such as the depth of a 
form-board, or 6". 

This line makes it possible, by mere inspection of known factors, 
to determine at once whether concrete stress should be investigated. 

To illustrate the saving of work accomplished by this, a case will be 
assumed where the intersection comes above the line, and depth is, 
therefore, to be determined by concrete stress : 

Column load 150,000. Soil value 1500. B/L = A6. 



GENERAL INFORMATION 



37 



Y 



150,000 
1500 



= 10' 



Shear depth, CKL= (1.00) (.0425) (10) =.425. 

As d c must be greater than this, .6 will be assumed temporarily. 

Then o = 4.60 + 1.20 + (10 — 5.80)/2 = 7.9. 

Bending moment from diagram, 

M = (.287) (150,000 X 10/10) =43,000 
Then by Formula (14) 



dl 



2 x 43 ■ OOP 



(3/8 x 7/8) (7.9) (144) (650) 



353 



-i 



353= .595 



It will be seen that the work involved is much more troublesome 
than the method using shear as the governing factor. 

One of the troublesome elements is the fact that d is a factor in the 
determination of width b, which in turn is used to determine d. This 
makes it necessary to cut and try. 

Bending Moments. For comparison different formulae in general 
use by competent, responsible designers have been applied to a footing 
having a common ratio of B/L = .42, selected at random. 







1 






-A2L , 




'///,. 


L 












H 






"?*< 



\ 



Alta Considered .29 £ .206L 2 .206 L 2 .206 L5 .14-4- L l 

Lever Arm .145 L .174. L .157L .\A-sL .202 L 

Product" M .042 1! W .02>54-L*w. Q3ZS>1?«, .OZ9qI a u>. .0291 LV 

Since Lfw = (plumnLoad * W thi& may be expressed reciprocal/ 1 <y as 
WL JVL WL WL vyl 

3/ 33.4- 35" 



M 



Fig. 



z5T3" 

Illustrating Bending Moment Assumptions. 



It will be seen that, while the assumptions as to area considered 
vary as much as one hundred per cent., the resulting distance between 
the critical point at the face of the cap and the center of action 
assumed for the load tends to lessen this discrepancy. Nevertheless, 
there is a difference between extremes in the final result equal to 
fifty per cent. 



38 HICKS — REINFORCED CONCRETE FOOTINGS 

It is also proper to say that more footings are probably in use. 
designed on the basis of the smallest value, representing the practice 
of one of the largest manufacturers of reinforcing products, than is 
true of any other type illustrated. The assumption in this formula, 
that the critical point for maximum bending stresses in the steel is 
on the line of the base of a frustrum of a pyramid, where rupture in 
diagonal tension would occur, appears entirely reasonable. The same 
reasoning is used by Talbot as the basis of selecting a point for the 
measurement of diagonal tension. 

Inasmuch, however, as there is an experimental basis for the Talbot 
formula, lacking in the others, it has been used in these diagrams 
throughout. Anyone desiring to use other values can do so by affecting 
the value of A s found from diagram with a coefficient representing the 
divergence of his formula from the one used in terms of a constant 
percentage, more or less than A s . 

Resisting Moments. The resisting moment in the steel has been 
used in the diagrams, or 

M = A s f s jd 

Equating this with the bending moment and solving for steel area, 
we have : 

t e 3d 

Where steel is uniformly distributed both ways through a square 
footing, the stress in the outside bars is less than in those near 
the center. 

Talbot concludes that the average stress of all the bars will be 
equivalent to considering that the bars within a proportionate length b, 
of the footing width, are equally effective. 

He defines this distance as: 

b= B4 gd + 1* - (B + 2d) 
2 

This length, therefore, defines the width within which steel for the 
bending moment should be distributed, and also the available area for 
calculating the fiber stress in the concrete. 

This distance b, in the notation used by Professor Talbot, is desig- 
nated in this book as M, or the width within which steel for bending 
moment should be placed. 

Area of Steel. The steel required for bending moment within the 
width b is ascertained by equating the bending and resisting moments 
and solving for A s in the usual manner. 

A simple expression for steel area true for all sizes of footings for 
a single cap ratio has been found as 



A s - 1 



and platted as curve A. 



GENERAL INFORMATION 



39 



This is so easily remembered that it is almost as simple to use it as 
to refer to the diagrams. The effect of change in size of cap is shown 
by curve P, which gives the correction to be applied to curve A, for any 
width of cap. 

It is of comparative interest to note that 12" reinforced footings of 
same mix and size as the unreinforced footings tested failed in 
diagonal tension under loading from three to four times as great as 
those causing rupture in bending in the unreinforced type. 

The reinforcing of the concrete in direct tension was sufficient to 
overcome the first line of weakness, whether due to lack of homo- 
geneity generally or specific weakness at a particular place in an 
example under test, until failure resulted from secondary stresses in 
the manner illustrated in Figure 7 as diagonal tension. 

Disposition of Tension Steel. Nothing conclusive can be derived 
from the tests as to the most advantageous position for a given weight 
of steel. 




Tension end bond failure. 




Fig. 9. 
Illustrating Failure by Ten- 
sion — Center Beam and 
Four Way Distribution. 
(Talbot.) 





,1 


cf: 


























i v> 


WorKi'riq Load 
























Z_ 


/ 




SZ500. 












/ 


ft 








Z7— 






Proper sire For 


























/ 




bars -s/a "a 






hf 






1 




















JV 




<■ / 














—I 


v 




_j 










*. 






/ 










:# 






zt. 










^ 






TL 










_T 






tz 










ZE 
















T 
















± 




a 












£ 




A i 












t 




7Z 












tl 




T^ 










































































.0 
— wo 


o 
9 


O C 


o <3 


o 

s 







J'* 


16 


3 n 


Ibx. 


be 


rn 


i 


**" Tifr-iA-U * 


V 4 


A 


A 




u j 




oL 



3/4 "ct> Max bond 206 lbs. Tens. 14900. 




Fig. 10. 
Typical Cracking by 
Failure. (Talbot.) 



Bond 



4-0 

HICKS— REINFORCED CONCRETE FOOTINGS 

The results shown by the four way reinforcement Ho w „■ 
UF& e v» ie T ° Ver *? T ay PlaeemenrNo^S r" nCedT 

wi^SSS3^"& rf '^2r when steel was placed 

be increased when required by ordinance conations 

the^d^^Z-Sotriro^/oS ^ ^ ""* betW6en 

centages of reinforcing avtfge values o'l Sf ^ Simikr pel " 
ing scale are secured as follow!? ert,C&1 shear m an ascend " 

003% .004% .005% .006% 

120 lbs. 124 lbs. 1361bs. l 6 01bs. 

a^^^S**^*^ i so inter - related t0 

inch value as a characteristtc of Z^™^ * ^^ a Square 

«&si^iftsssrri^ after the change ° f «~ 

the steel passes its elaTticTtnit or f^*^ t ? nS '° n resistan <^ when 
bond is reached ' ' Cntleal po,nt of the concrete 

other types of weakness do not become cri t "c a l SU ° h **** that 



GENERAL INFORMATION 



41 



The following data has been selected from page 104 as presenting 
failures by diagonal tension, most clearly distinguished from other 
types of weakness. 



No. 



Con- Depth Depth 
Over to 

All. Steel. 



crete. 



Bars. 



1447 1-21/2-5 12 

1448 l-2V 2 -5 12 
1552 1-21/2-5 12 
1808 1-2 -4 12 
1810 1-2 -4 12 



Per- Value 
centage. of J. 



10 11 pi. 0.31 

10 11 pi. 0.31 

10 10 1/2 sq. cor. 0.42 

10 27 % P l. o.5D 

10 33 % Pl 0.61 



5 %,2 1/2, 
1435 1-21/2-5 II1/2 10 2 1/3 & 2 % 0.61 



1554 1-21/2-5 12 



.90 

.90 

.89 

.895 

.89 

.87 



cor. sq. 
10 15 1/2 cor. sq. 0.62 .87 



Disposition. 



6 par. to 
each s, 

5 par. to 
each d 

Do. 



6" C. to C. 
in 2 way 

2%" C. to C. 

2 way 

lit" C. to C. 
2 way 

5" C. to C. 

2 way 



4" C. to C. 

2 way 



46/60ths of steel considered in figuring percentage of reinforcement, 
and 46/60 of area of concrete above steel for same purpose. 



Calculated Stresses. 



No. 



Load at Steel 



First 



Vert Bond 
Failure Tension at lc j Stress Cracks 



Others 



1447 208000 42100 130 



1448 176000 35600 110 



1552 236000 45100 149 415 



1808 198000 31600 218 218 



1810 219000 28700 198 198 



1435 208000 22000 134 



1554 288000 37800 185 346 



144000 160000 167 

center N face g. face 

^ °1 center center 

Vv . face 

14400& 



Failure 



Sudden 



3 faces 




Sudden 


center 






102000 
3 faces 




Violent 


center 






120000 
N. face 
center 


138000 
N.E. 
& W. 

center 


Diag. 

Tens. 


2 faces 
center 




Diag. 
Tens. 


136000 
W. face 
center 


144000 
E. face 
center 


Diag. 
Tens. 


120000 
center 




Diag. 
Tens. 



42 



HICKS — REINFORCED CONCRETE FOOTINGS 



No web reinforcement of any kind was present in any footings 
tested. 

All footings five (5) ft. square with a cap 1 ft. square. 

In order to relate the stresses calculated from these experiments to 
working design, the working load, for which the dimensions used are 
adapted, will be calculated on the assumption that limit shear at 
120 lbs. is the governing factor. 

Then external shear V= (4) (12) (10) (%) (120) =50400. 
Column load (25/24) (50400) = 52500 lbs. 
Soil load 52500/25 = 2100 lbs. sq. ft. 

The reinforcement used bears no direct relation to any design load, 
as its amount was varied to observe resulting changes, but the dimen- 
sions used for all the footings are adapted to working soil reaction 
of 2100 lbs. per square foot for maximum limit shear of 120 lbs. at the 
face of cap. The soil pressures created by test leads, and the unit 
vertical shear at face of cap at the time of first crack and at ultimate 
loads, have been calculated as follows : 





1st 


-SOIL LOADS— 




UNIT SHEAR AT FACE OF CAP 

1st 


No. 


Crack 


Ultimate 


Design 


Crack 


Ultimate 


Working 


1447 


5700 


8300 


2100 


340 


490 


120 


1448 


5700 


7000 


2100 


340 


415 


120 


1552 


4000 


9500 


2100 


237 


550 


120 


1808 


4800 


7900 


2100 


281 


466 


120 


1810 


5500 


8800 


2100 


322 


515 


120 


1435 


5400 


8300 


2100 


308 


474 


120 


1554 


4800 


11500 


2100 


274 


655 


120 



It is apparent from the above tabulation, that a factor of 2 against 
first sign oi* critical weakness in diagonal tension existed in the 
poorest specimen without special reinforcement. 

Compare the bond stress in two typical examples, as, tor instance, 
No. 1552 and No. L810. Assume bars 56" long. 

No. 1552. Perimeter 2 2~q - 20 



GENERAL INFORMATION 48 

Effective (46/60) (20)=15.35. Effective length (56 — 12)/2=22". 

Then (15.35) (22) =338 sq. in. bond area available. 

Tensile stress @ (45100) (10) (46/60) (.25) =86600. 

Unit bond resistance developed, 86600/338 = 256 lbs. 

No. 1810. Perimeter 1.18 2 Q = (3 3) (1 . 1 8) = 3 9 . 

Effective (46/60) (39) =30". 

Then (30) (22) = 660 sq. in. available bond area. 

Tensile stress @ (28700) (33) (46/60) (.11) =80,000. 

Unit bond resistance developed, 80,000/660 = 121 lbs. 

The results are consistent with the idea that the low diagonal 

tension failure was associated with a relatively high bond and tensile 
stress. 

The significance of the cracks should also be considered. 

The cracks were practically all tension cracks generally located in 
line Avith one face of the cap, and the final fracture is typical in plan 
with those observed in unreinforced footings, representing beam failure 
in bending. See Fig. 6- 

It appears that first tension cracks do not penetrate beyond the 
rupture plane for diagonal tension, since the central frustrum is 
intact with cap after failure. 

Also the breaking away of the reds from the concrete, outside the 
frustrum constituting bond failure, finally destroys the tension resist- 
ance after diagonal tension cracks are formed, and permits the sudden 
separation in diagonal tension and the breaking by beam action as in 
plain concrete. 

This indicates that low bond values for working stresses and rein- 



44 HICKS REINFORCED CONCRETE FOOTINGS 

forcement against incipient diagonal tension will generally increase 
the factor of safety of a footing. 

Tension cracks weaken bond resistance, and ultimately bond weak- 
ness permits final failure by diagonal tension. It is practically impos- 
sible to disassociate these factors. 

It should be remembered that ultimate failure developed resistance 
from one and a half to two times greater than that shown at first 
crack. 

Talbot affirms in general that web reinforcement takes no stress 
until cracking occurs, but as incipient diagonal tension cracks must 
exist internally long before tension or bond weakness becomes critical, 
the presence of reinforcement normal to the plane of rupture would 
evidently assist in preventing failure from this cause. 

It appears from the facts noted that while reinforcement for 
diagonal tension might not be needed if concrete were as reliable as 
steel, yet it should be used anyway to guard against locally defective 
construction, peculiar to the material. 

Reviewing the results presented by No. 1552 : 

Shear at 
Bond face of Shear 

Tension Max. Average cap at Id 

Load at first crack 102000 19500 

Load at ultimate 236000 45100 

Load at working limits 52500 10200 

A diagram showing the lateral distribution of vertical shear and 
bond from the face of cap to the outer edge of footing 1817 is shown in 
Figure 11. The bond curve takes the ordinary parabolic form, while 
the curve of the shear is an inverted parabola, similar to the bending 
moment curve. 



179 


111 


237 


64 


415 


256 


550 


149 


92.5 


57 


120 


33 



GENERAL INFORMATION 



45 



< DiT d 10 



<v < 



t-4. isc) cop i i s« 
to- fi V* 7k*C iwo^ 
p-o-4i Uiricod «59ooc.Tcn iofecc 

Ayr wcrKino \xods W ZJOC V» j4ccc 
;v Itc id A2 U» 




£50 



.3SC 



Fig. 11. Ilixstbating DisTBiBrTiox of Bond and Yebtical Sheab- 
FoornsG 1817 (Talbot). Fajxuee by Bond. 



Local ordinances fix position at which vertical shear in exe— 
40 lbs. allowed in the concrete is to be used as an index for diagonal 
tension. 

This position is usually expressed as Id from face of cap. jd. d 2. or 
some other fraction of depth. 

In the case of footings under test, the point selected at which to 
measure vertical shear as an index of diagonal tension is beyond the 
point of 40 lbs. shear at working limit of of 120 lbs. maximum shear 
at face of cap. See Fig. 11. 

Many designers accept this as warrant for omitting web reinforce- 
ment altogether. This is not only contrary to the spirit of the 1913 
Joint Committee report allowing 120 lbs. shear at face of cap. when 
all tension normal to plane of shearing is provided for by reinforce- 
ment, but involves a distinctly inadequate factor of safety for a 
material of known uncertainty of deportment. 

Other designers ascertain the unit shear at the point selected, and 
if in ex: sag E40 lbs., reinforce for two-thirds of the stress in the steel, 
but provide nothing, if less than 40 lbs. For 42 lbs. unit shear they 



46 HICKS — REINFORCED CONCRETE FOOTINGS 

provide steel for 28 lbs., but if the unit shear is 39 lbs. omit reinforce- 
ment entirely. 

Others again provide steel for two-thirds of the unit stress and 
presumably, where the point fixed is beyond the point of 40 lbs. shear, 
assume some related distance to use in calculating the sum of the shear. 

The range of application varies from nothing to the provision of 
absurdly large amounts of web steel. 

The greater number of footings are probably constructed for soil 
pressures ranging from 4000 to 8000 lbs. per square foot, and with 
ratios between B and L ranging from .25 to .45. Illustrating the 
range of values of the factors of d/L: 

Soil— 4000 5000 6000 7000 8000 

K 11 .14 .17 .20 .23 

B/L 25 ,30 .35 .40 .45 

C 2.08 1.77 1.48 1.22 1.04 

It will be seen that the minimum values of products of these factors 
would occur with low soil values and high cap ratios. But low soil 
values mean large footings and relatively low cap ratios. Take an 
actual case of a hooped column 24" square to carry 400,000 lbs. at 
4000 lbs. soil reaction, This would require a footing 10 ft. square and 
a cap 3,33 ft. square at most, or 

B/L = 3,33/10 = ,333. 

TKejiCK = d/L= (1.56) (.11) =0.17, or 

d = 0.17 L. 

It will be seen from examination of the curve of S on diagrams that 
the distance from the face of cap to point of 40 lbs. shear is as follows : 

B/L .25 .30 .35 .40 .45 
Side of square for 

40 lbs. shear. . . .56 L .615 L .67 L .71 L .75 L 

(S/L — B/L)/2.. .155 L .1575 L .16 L .155 £ .15 L 

The values in line 2 represent the sides of a square, where 40 lbs. 
shear exists in footings having B/L as tabulated. Then deducting 
width of cap and dividing by 2 will give the distance out from face 
of cap to point of 40 lbs. shear as a fraction of L. It will be seen that 
within the limits considered values of this distance range from .15Z 
to .16L. 

As the least probable depth for footings within same limits was 
found to be AIL, it is evident that, with but few exceptions, the point 
id from face of cap will be beyond the vertical plane where vertical 
shear of 40 lbs. exists. 



GENERAL INFORMATION 47 

In the ease of the footings tested by Prof. Talbot it has already 
been shown by Figure 11 that only 30 lbs. of shear existed at working 
limits Id from face of cap. 

This may be cheeked from diagram : 

For B 1 = 1 5 = .2, C = 2.-4. 

For working soil value 2100 lbs.. K=M 

<K= 2.-4 .059 =.167 = d L. 

Then d = . 1671. or. .167 5 12" =10". 

For point of 40 lb. shear read across en B L = .2. to curve of 8, 
and find value on Lower Scale of AloL. 

.475 — .21 2 = .13751= .1375 (60) =8.25". 

It will be noticed that the position of the point of 40 lbs. shear at 
working loads was about one-third the width of the projection of 24" 
beyond the face of the cap. and as the effective depth of the footing 
was 10". the previous conclusion that the point Id would be beyond 
the point of 40 lbs. shear is verified, even with such a low soil value 
as 2100 lbs. 

It can also be shown that the use of 1 - 2 d is inconsistent, since it 
must be assumed that the purpose of the ordinance in providing a 
method for measuring diagonal tension is to require some minimum 
amount of effective web reinforcement in all cases. 

It will be seen again that, since the distance of the point of 40 lbs. 
shear is from .15Z to .16L from the face of cap. W7 will be at or 
beyond this point in all footings having a value of d L of .32 or over. 

Examination of the factors of d L between soil values of 4000 and 
- lbs. show that a soil value - I lbs. and a B L of .35 give 
d/L= .34. or 1 2 d is slightly beyond the point of 40 lbs. shear. 

The same result would follow if the soil pressure were 6500 lbs. 
with a value of K of .184. and a B L of .30 having a value of C = 1.77. 

The product of these two factors is 1.80 .1-4 = .33. and one-half 
of this amount evidently coincides closely with the point of 40 lbs. 
shear which has been shown to be .1575Z from face of cap. 

It may be accepted as true that many Building Departments will 
continue to permit the omission of web reinforcement under such 
circumstances until ordinances are changed, but it is well for respons- 
ible architects and designers to know that, at best, they are dependent 
for a somewhat uncertain factor of safety of about 2 upon the mix 
and placement of their foundations being equal to thos - ired in 
laboratory experiments. 

Pending the readjustment of official standards, conservative 
designers will provide what they may regard as a reasonable minimum 



48 HICKS — REINFORCED CONCRETE FOOTING? 

regardless of the fact that a slight saving may be made by misusing 
official sanction. 

It should be understood that the steel used is to insure the validity 
of the factors of safety of about 2 already existing in the concrete as 
against uncertainties of workmanship and inspection. 

If a local pocket of loose aggregate exists in the web by reason of 
carelessness, the resulting tension crack will induce secondary stresses, 
which would ordinarily be within the capacity of the concrete to 
resist, but, because of the local pocketing and incipient diagonal 
tension cracks, these stresses are transferred to the steel crossing the 
plane of rupture. It is evident that such a pocket would be localized 
even in the worst work, and could only be a fraction of total effective 
resistance afforded by the concrete itself. Using an average value of 
B/L = .4, the sum of the entire vertical shear from the face of the 
cap to the point of 40 lbs. shear is: 

Formula (9). 

v= (80/2) (.706 + .40)/2) (.155; (144;Z 2 = 494Z-'. 

The actual sum is somewhat less than this, because the bounding 
line of the shear curve is parabolic in form and the assumption of a 
triangular distribution, substantially correct for small values, would 
probably involve an over-estimate of 10% with such a large value 
as 80 lbs. 

If this result were applied to a footing 10 feet square with a soil 
pressure of 6000 lbs. per square foot and a depth of C K L = 
(1.22) (.17) (10) =2.07 ft., the area of bent steel in each direction 
would be 

(0.7) (494L 2 )/16000, or 2.14 sq. ins. 

It is hardly conceivable that a local irregularity of placement or 
mix could extend over 10% of the section, because, even assuming 
the existence of a pocket where the diagonal tension stress is maximum, 
one cannot practically imagine the reduction of effective section to 
such an extent. Also, it is a fact that there would be no actual pocket, 
but only a reduction of the effective resistance of the defective portion 
of the concrete. 

As the steel commences to take stress in proportion to the weakness 
of the concrete, it seems probable from analogous experience with the 
deportment of tension steel in the presence of web weakness, that the 
deficit in strength would be distributed through all the steel crossing 
the section of weakness. The direct tension steel would be 

As— (.98) (7.15) = 7 sq. ins. 

The proportion between the tension steel and that for diagonal 
tension would then be 2.14 7 =31%. 



GENERAL INFORMATION 49 

If 10 c c of the total sum of vertical shear is used as a measure of 
minimum reinforcement, where ordinance permits the omission of 
web reinforcement entirely, there would be required each way .214 
sq. in. or area equivalent to 3.1 c / c of the tension steel. 

It will be found that this is practically equivalent to the result that 
would follow selecting the place where 60 lbs. of vertical shear exists 
as the point to measure vertical shear as an index of diagonal tension, 
as follows : 

60-40 .706 6.06 (.05) (144) L* 

Ays= — 3 = .296 sq. ins. 

vs 16000 H 

or 3^ of the tension steel. 

It will be seen from the form of the last equation that if the value 
of v were a known constant, fixed by competent authority, as for 
instance 60 — 40 lbs. as a limit to define the minimum of reinforce- 
ment, that the equation would become : 



(20) (.05) (144) ( Z + T^ 



At s = 16000 ; L ' or 

(.0045) (| 4 {l) (if) 
A v 8 ~ 2 

or in the above case, (0.7) (.0045) (.706 + ,606)/2(100) = .21 sq. in. 

This is possible because the value of y in Equation (7) for any 
given value of v is practically a constant throughout the range of B/L 
in use and may, therefore, be substituted by its average numerical 
value, which for 20 lbs. in excess of 40 lbs. allowed in concrete is .05. 

It will be seen that the last equation is identical in form with 
general formula (7). 

The values of 8/L and 8 t /L are obtained by inspection from the 
diagram for B/L used, and the minimum area of steel varies with L 2 , 
or is as the area of the footing. 

Any argument or contention that the point of 60 lbs. shear is a 
more suitable point than another would be valueless, but no one is apt 
to contend that less than the amount of steel required by this selection 
should be used. The limiting extremes have been pointed out and any 
intermediate point required by local authorities may be used with 
comparative knowledge of its import. 



50 



HICKS— REINFORCED CONCRETE FOOTINGS 



As will be shown later in the discussion of bond, from fifteen to 
twenty-five per cent, of the tension steel can be bent up without any 
significant increase of expense other than for bending labor. 

The amount required for diagonal tension on the basis of designing 
for 20 lbs. of vertical shear in the steel will be from 4% to 5% for 
stirrups and from 3% to 3y 2 % for bent bars. 

It follows, therefore, that for practical designing purposes the 
intelligent disposition of the material available will probably cover the 
local requirements anywhere. 

If vertical stirrups are used, the continuous type made of %" plain 
round or smaller sizes insures satisfactory bond conditions without 
calculations, secures thorough distribution of web resistance and 
involves practically negligible expense. 




5/8" sq, 
'/A" <£ 



def. bare. 2. wcr/6, shraiqhf. £ 
plain conHnotfus sHrrups v> 




Fig. 12. 



Illustrating Use of Continuous Stirrups for Shallow 
Footings in Place of Bent Bars. 



Suppose that in the footing just used for illustration the tension 
steel was provided by 18 % square deformed bars in each direction 
for widtli .1/ (See Fig. 12) — 

M- 4+4. 14 + 10 T 8 ' 14 = 9.07 ft. 
2 






GENERAL INFORMATION 51 

Then width requires 

. 10 - 9.07 . 

/ 3\(__2____ ) (18) - .62 bars 
\tr io 

As eaeh bar is .39 sq. in. area for 0= (.39) (.62) = .24 requiring 
1 y 2 " square bar. 

There are then twenty bars to be placed in the width of the footing. 
Absolutely accurate spacing laterally and vertically may be secured 
by using 4 continuous stirrups with as many double legs as there are 
tension bars. 

The tension bars are suspended in their correct relative position 
with no ties required between the stirrups and the tension steel. 
The amount of steel required will be roughly — 
(20)(2)(2'0") =80 ft., or 
80 ft. @ .167 lbs. = 13.35 lbs. 

13.35 lbs. @ 2f« base 26.7^ 

13.35 lbs. Standard extra @ .50 . . . 6.7$ 

Bending 4.0^ 

Placing 2.0^ 

Total cost per stirrup .... 39.4^' 

The figures for bending and placing are made up from actual cost 
figures. The wire is wound around pegs on the shop table in form of 
a figure 8, is sent to the job in that shape, and pulled out to space and 
length when placed. Seventeen double legs are effective in width M 
as diagonal tension reinforcement, or 

(34) (.049) == 1.67 sq. ins. = 23.7% 
of area of tension steel. 

At first glance this arrangement seems objectionable, but study of 
its advantages will show many strong points. The ends of the con- 
tinuous stirrups are extended up and hooked over the edge of the 
forms, and the stirrup is supported at intermediate points on frag- 
ments of brick, making a strong, reliable support for the tension steel. 

When considered as a percentage of the cost of tension steel, cost 
will be seen to be a negligible factor of expense. The cost of tension 
steel would be : 

36 bars % sq. @ 1.33 X 9.67 465 lbs. 

4 bars y 2 sq. @ .85 X 9.67 33 

498 lbs. 

498 lbs. @ 2^ base $ 9.96 

Placing @ y 2 f 2.49 

Total cost $12.45 



52 HICKS — REINFORCED CONCRETE FOOTINGS 

Cost of stirrups, 1.57/14.02 = 11% of total steel cost. 

In the same way it might be shown that the stirrups represent less 
than 3% of the cost of the entire footing. This might be slightly 
reduced by using steel proportioned to the actual stress, but as it 
represents the practical minimum for efficient handling and is evi- 
dently much more convenient than a multiplicity of separate stirrups, 
it makes a convenient element in standardized designing. 

It further illustrates the fact that the design will be independent 
of the exact point selected at which to measure diagonal tension. 

The position of web reinforcement will be considered in connection 
with the subject of bent bars, but the general conclusions may be 
summarized as follows : 

Minimum amount of reinforcement will be determined by position 
of measurement fixed by ordinance or interpreting authority. 

Maximum amount will be fixed by the designer through the use of — 

1st. A percentage of tension steel bent up. 

2d. Continuous vertical stirrups. 

3d. A combination of 1 and 2. 

The amount of diagonal tension steel required may vary from 
nothing to as much as 40% of the tension steel. 

From 15% to 30% of tension steel may be bent up without affecting 
working bond limits for remaining steel, with negligible expense. 

Regardless of requirements, not less than 10% of tension steel 
should be bent up, or equivalent resistance in the form of vertical 
stirrups provided. 

Continuous stirrups representing from 20% to 50% of tension 
steel area involve negligible expense, and for important work may be 
combined with bent steel. 

Conclusions reached by Professor Talbot on beam experiments 
leading to the recommendation that both bent bars and stirrups are 
desirable, seem applicable to the conditions presented in footing design, 
both from the standpoint of assured stability and of negligible cost. 

Disposition of Web Steel. While the vertical shear is maximum on 
the line of the edge of cap, it has been shown that rupture by diagonal 
tension occurs before punching shear becomes critical; also, that the 
tension cracks preceding diagonal tension failure do not extend across 
the plane of rupture by diagonal tension, and that the frustrum of 
a pyramid produced by the planes of rupture by diagonal tension is 
integrally intact with the cap after rupture. 



GENERAL INFORMATION 53 

It follows that no diagonal steel will be needed across the vertical 
plane in line with the cap, but that as the decrease of bending moment 
permits tension steel may be bent up at an angle of 45°. Its ideal 
position from the consideration of symmetry of section would be such, 
that it would cross the center of the plane of rupture, which would 
be somewhat below the geometrical center of the section above the steel. 

The clearance of bent up steel below the upper surface of concrete 
will be from 2 to 3 inches, and the point of bending will ordinarily be 
from 3" to 8" out from the face of the cap. 

This means that its intersection with the plane of rupture will be 
somewhat below the point indicated, but as the diagonal tension cracks 
start at the bottom of the footing, a compensating advantage is gained 
through the steel crossing the rupture plane lower down, and also an 
increase of available anchorage. 

When excess steel beyond requirements is being considered, any 
necessary addition to grip length may be obtained by decreasing the 
angle of 45° instead of making a second bend above. 

The area of steel required will be increased as the sine of the angle 
between a vertical from the point of bending and the inclination of 
the bar. 

Where in shallow footings the point of bending is so close to the 
plane of rupture as not to afford a satisfactory position, continuous 
stirrups may also be employed. 

When used these should be located at the point where a vertical 
line intersects the plane of rupture half the length of the stirrup 
above the tension steel. This position insures an equality of anchorage 
above and below to the maximum possible extent, a consideration of 
much more importance than its location in the center of gravity of 
the shear area used as an index. 

The location of this index merely establishes the magnitude of the 
assumed stress, and does not determine the most advantageous position 
for the steel to resist it. 

Calling the clearance for the steel below the upper surface of the 
concrete C, the advantageous position of continuous stirrups will be 
(d — c)/2 beyond the face of cap. 



'A 



HICKS — REINFORCED CONCRETE FOOTINGS 



„_^J__L ^_1_^. IJzL 




Sttnt £>orj available 
Be^^^ Bars required maxi 



Continuous Kerr.STi'rrufS mat. 

Use «»" o def bars £4 *»oi . 

Then ^6 stnqle leqs 5/«i"e) eh'rrup&O .osa 

Wqr of 4 CofiHnVOVS shrrgfU 






25k 



^g±^ 



\ 



.•^.j.^.-_lL^.l.^j 






H'9' 



irajr' 



15 ^4'odtf. 2~o 




H-+-H- 

H-H4 
B-+-H 



=f=R 



ii 



4-H--H 



=ra 



T 

-r 



4-H-H~H-h-i-f 



.—1—1—1-4 



I 37 'It'odrf, a ban r up. Z»q 



bbbbii^ 



Ins. 'I 



wMm 




B/L 


.30 


.33 


40 


45 


.50 


.5 5 


.60 


•S3 


.67 


7i 


76 


H 


364 


3.54 


3.64 


3.54 


3.54- 


3.34. 


3.54 


354 


354 


354 


3.34 


1 

d 

Cap siie 

CopConcrete 
6ose&>ocrete 


.oo 


.56 


MO 


1.46 


1-76 


2.06 


Z.30 


242 


2.56 


2.76 


2.9/ 


3.64 


2.96 


2.44 


2.06 


1.76 


1.46 


1. 24 


112 


.96 


.76 


.63 


O.OO 


4o8" 


4.65 


523 


5.81 


6.40 


699 


733 


7.80 


35o 


663 


.oo 

482. 


IO. 
4o3 


Z4. 
332. 


40. 
Z63. 


60. 
24o. 


35. 
20>. 


112. 
169. 


131. 
152. 


137. 
131. 


202 
I03. 


229. 
65. 


Concrete 


482- 


4 13. 


356. 


323. 


3oo. 


2.66. 


28/. 


263. 


266. 


305. 


514. 


•Sq.lnsSree/ 


833 

657 


9.(6 

I6Z5 
105.25 
II9.60 

2 /l 


9. fee 
74| 

•7o6 

asoo 


9.75 


9 73 


s.ez. 


a»e 


8.3 6 


638 


741 


661 


(P*Tof Steel 
(9»f&ncrete. 
Torol <gt,h 


* 15.10 
'20.5O 


'73© 
oO.TS 


753 
>73© 
7S.oo 


742 
17.07 
7I.SO 


70o 

16.23 
7025 


664- 
15.70 
70.7S 


646 

M-65 
72.oo 


572 
13. IS 
7G.Z5 


526 
IZ.IO 
76.5o 


Co f Depth 
Cop projechon 


o/o 


106X16 

.. 9/| 


98o5 
-5/3 


9z.5o 


68-57 
1.46/, 


(36.50 
4/3- 


66 A 

s/4- 


6635 


&9Ao 

"/io 


9oeo 
,2 /.2 



Fig. 13. Illustrating Economic Design. 

Bond Stress. The usual method of calculating bond is to consider 
satisfactory, if the product of the bar perimeter, the length of bar 
beyond the point of maximum tensile stress, and the allowable working 
bond stress is equal to, or greater than, the product of the bar area 
and its allowable tensile stress. 

Working bond stress is, therefore, the average stress to which the 
concrete along a reinforcing bar is subjected per square inch of contact. 



Tli is is also the meaning attached to the term in building ordinances. 



GENERAL INFORMATION 55 

A method of selecting a proper size of bars for known conditions 
to develop exactly allowable limit bond stress has been devised that 
permits this result to be attained without calculations by a mere 
inspection of the bond curves. Each curve is designated on the 
diagram by the size of the bar above and its perimeter below the line. 

The curves are points of equal bond stress for bars of given size. 
Since the value of the resistance for a given bar is as the surface area 
of a unit length, the variation in resistance between bars of different 
size will be directly as their perimeters. But the variation of tensile 
strength is directly as the areas, or as 

. St 

4/ - 16 

It is then evident that if a bar, having a perimeter of 4, furnishing 
(under the conditions of its placement) equal resistance to tension 
and bond, is divided into four bars under the same conditions, that 
while the combined tension resistance of the four bars remains the 
same, the bond resistance is doubled, or is inversely as the perimeters. 
Illustrating : 

Perimeter of given bar 4. 

Area of given bar 

16 16 

Area of each of 4 bars . 0.25 

Perimeter of divided bars 2. 

Sum of divided perimeters 8. 

Katio of perimeters 4/8 

Since there are four bars, it is then evident that two of them 
provide resistance to bond stress equivalent to the tensional value of 
all. As this merely represents the law of arithmetical relation between 
area and perimeter, it is true for all perimeters, and a general rule, 
availed of in the use of these diagrams, may be stated, viz. : The 
amount of steel of a given size, required to develop bond resistance 
equivalent to the tension value of all the tension steel at the critical 
point, may be found by multiplying the area of tension steel required 
for maximum bending moment by a fraction, whose denominator is 
the perimeter of a bar that would provide equality of resistance to 
working stresses in tension and bond, and whose numerator is the 
perimeter of the bar to be used. The difference between the amount 
so determined and the amount required for maximum bending moment 
represents the area that may be bent up, whenever the decrease of 
bending moment permits. 



56 HICKS REINFORCED CONCRETE FOOTINGS 

In using this simple fact with the diagrams, it will be seen that, 
while the distance between the curves measured on vertical lines 
varies, the value of the separation is uniformly one-half. That is, the 
lines are contours of equal value one-half unit apart. 

If absolutely exact determination of the values is desired, the 
co-ordinate spaces, between the curves, may be counted and the posi- 
tion of the intersection of B/L and L precisely located on this scale, 
but the eye will easily assign the position of this point a proportional 
value of sufficient accuracy for the purpose in hand. 

Illustrating by the use of data in Example 1 : 

The intersection of B/L and L is about 1/5 (actually 1.8/9) below 
the %" curve, which has a perimeter value of 3. Then this fraction 
represents 1/5 of one-half (the increase of perimeter value in the 
distance to the next curve below), or 

(i/5)(y 2 ) = .i 

Then 3.1 represents the perimeter of a bar that would furnish exactly 
equal resistance in tension and bond, under the stated conditions of 
placement. 

Checking this conclusion : 

The projection of footing, (10 — 4.5)/2 = 33 inches 
Clearance for steel at face of footing . . 2 

Length of bar available for bond resistance 31 inches 
Then bond resistance = (3.1) (31) (100) = 9610 lbs. 
Area of bar 

3.1* . 9.61 



4 16 

As tension in steel is 16000 lbs., the tension value of this area is — 
(16000) (9.61 )/16 = 9610 lbs. 

Now, as some steel is to be bent up, and as excess will be beneficial 
so long as the area is sufficient for the estimated diagonal tension 
without diminishing the allowable resistance in bond and tension, 
% bars having a perimeter value of 2% will be used. Then the 
number of bars required in tension will be — 

7.13/.391 •. . =18.30 

Number of bars required in bond, (2.5/3.1) (18.3) =14.75 
Number of bars available for bending up . . 3.55 

Checking these conclusions : 

Total tension, (18.3) (.391) (16000) = 114000. 
Total bond, (14.75) (2.5) (31) (100) = 114500. 



GENERAL INFORMATION 57 

The inaccuracy indicated in the check of about four-tenths of one 
per cent, is, of course, negligible and is probably due to error of 
position in platting of curves. 

The simplicity and precision of the method used contrasts favorably 
with the usual custom of guessing at the size of bars, and then 
accepting the result as 0. K. because needless excess strength is found 
to be available. 

The tension value of any bar of known perimeter can be figured 
for working stresses of 100 lbs. and 16000 lbs. respectively in bond 
and tension without determining the area, by the formula : 

Tension value, (1000) (O 2 ) for square bars. 

Tension value, (1273) (O 2 ) for round bars. 

Also the greater relative efficiency of square bars as regards bond 
may be illustrated as follows : 

Equal Equal Equal 

Diam. Perim. Area 

Kound Square Square Square 

Diameter 1.00 1.00 .7854 .886 

Perimeter 3.1416 4.00 3.1416 3.544 

Area 0.7854 1.00 .6169 .7854 

It thus appears that, for bars of same area, square bars give greater 
bond resistance than round ones. 

Also, it may be stated that while many bond failures with plain 
bars were noted in the footings tested, only one failure occurred with 
deformed bars, and this was at the relatively high unit stress of 596 
lbs. per square inch. 

One of the normal failures with plain bars was at a unit stress as 
low as 223 lbs. The relations will be seen in tabulated form more 
readily. 

Working Stresses : Ultimate Factor of 

Average Maximum Failure Safety 

Plain bars 80 130 223 1.72 

Deformed bars 100 163 596 3.66 

From "pulling out" tests at the University of Illinois, the following 
was obtained : 

Plain Johnson 

Lbs. Lbs. 

Maximum 360 639 

Minimum 174 298 

Average 281 484 



58 HICKS — REINFORCED CONCRETE FOOTINGS 

It should be emphasized that full size working tests do not usually 
realize the unit values derived from "pulling out" and "pushing out" 
tests. Such tests, however, appear to establish a ratio between plain 
and deformed sections that is reasonably dependable, viz., that the 
minimum values for deformed bars are relatively much higher than 
the low records for plain bars. It would appear from the observations 
here considered that the working stress for plain bars is too high for 
reasonable safety. 

Anyway it will be seen that for the working values used in the 
diagrams it is distinctly advantageous to use square deformed bars. 

It will also be useful to remember that other tests indicate that 
bond resistance is increased by the presence of vertical stirrups. 

The bond curves have been platted for this condition of maximum 
efficiency, viz., square deformed bars @ 100 lbs. working stress, to 
make the best way the easiest way. 

Other values of working stress, either larger or smaller, may be 
integrated into the results by multiplying the perimeter, affording 
equal resistance in tension and bond, by a fraction whose numerator 
is the stress to be used, and whose denominator is 100, the stress for 
which the curves are figured. 

The importance of bond has been appreciated by very few designers. 
The experiments of Prof. Talbot emphasize its importance and cause 
him to express a caution as to the consideration to be given it. 

It is analogous to providing sufficient rivets in a plate girder to 
develop the full working strength of the material required in the 
flanges and web to resist the external load stresses, and to such distri- 
bution of these that the greatest localized intensity of stress will not 
exceed the safe capacity of one rivet. 

In the same way, the measure of maximum tensile stress that can 
be developed in the reinforcing steel will be the safe capacity of the 
concrete in shear at the critical point to take off the decrement of 
stress, this resistance being measured as bond stress. 

It. therefore, becomes important to understand the distribution of 
bond stress, its maximum intensity, and relation to average bond stress 
as used in the meaning assigned to it in building ordinances. 

.Maximum intensity and distribution may be determined by Equa- 
tion (15) — 



U 



mojd 



;is shown in Figure 11, which represents the bond curve for one of the 
Talbot footings. 






GENERAL INFORMATION 59 

The maximum bond ordinate evidently occurs where the bending 
moment is maximum, and, since the entire external shear is used, 
assumes that the bars extend to the face of the footing. 

The bounding line of the bond ordinates is a parabola, and the 
value of its mean or average ordinate is, therefore, 2/3 of the 
maximum intensity. 

If this were a known constant relation, Formula (15) would be a 
useful one for design, as factors of safety should be definitely related 
to the maximum intensity of stress rather than to average values. 

Actual examples show that the relation is not uniformly constant. 

Until this relation is rationalized and ordinances are changed 
accordingly, it is necessary to continue the present method, but it is 
evidently desirable to investigate the relation as its exists in typical 
instances of design and in experiment. 

The following tabulations are from diagram calculations : 



Soil B/L d 


As 


Tension 
Capac. @ 

16,000 lbs. 


Bars 


w 

Available £'- J 
Size for Bond £g 


m 

r— O 

£2 


7000 .3 3.54 
7000 .6 1.24 


6.25 
6.72 


100,000 
107,500 


11.15 

48. 


% sq. 8.37 40 
3/ 8 sq. 33. 22 


o _ 
QQ© 

100,400 
108,900 


Then for 













»•* u = (e. a tyB?i7«ff&;a)(i») - w ' B 

Uo.2 (.16) (7QQ.0Q0) 17A. 

(35) (1.5) (7/8) (1.24) (12) 

In these cases the average bond is figured for bars having a clear- 
ance of 2" from face of concrete, while the maximum bond uses the 
entire external shear so that the results are not strictly comparable 
for other conditions than those stated. 

For use with the diagrams, however, it is evident that the use of 
100 lbs. working bond stress involves maximum stress at critical 
points of about 175 lbs. per square inch. 

The following is from Bulletin 67, University of Illinois : 











Tension 








to 

> ^ 


It 










Capac. @ 






Available 


.Sj 


M-< 


Soil 


B/L 


d 


As 


16,400 lbs. 


Bars 


Size 


for Bond 


0J c 

fcio 


PQ© 


1414 


.2 


0.83 


4.06 


66,300 
<S> 34,800 lbs. 


9.17 


%0 


9.17 


21 


66,300 

(a) 155 


1806 


.2 


0.83 


1.86 


64,500 


16.85 


% 


16.85 


21 


64,500 



CO HICKS REINFORCED CONCRETE FOOTINGS 

Then for No. 1418, u = 226. Eatio 226/146 = 1.55. 
No. 1806, u = 241. Ratio 241/155 = 1.56. 
If all the instances given were figured on the assumption that the 
bars extend to the face of the footing, the results would be : 

No. 1. (40/42) (100) = 95.5 Then 171.5/95.5 = 1.80 

2. (22/24) (100) = 91.6 174/91.6 =1.92 

1418. (21/24) (146) = 128, 226/128 = 1.765 

1806. (21/24) (155) =136. 241/136 =1.77 

The low values of bond failure given on page 96 of Bulletin 67 are : 

No. 1417. Eccentric loading 206 lbs. 

1451. Sloped footing 131 

1833. Outside bands 152 

1834. Outside bands 206 

1837. Center bands 198 

As the loading of 1417 was not uniform, the results are not com- 
parably determinable. The results in 1451 are clearly due to increase 
of bond stress, because of change of value of jd, through sloping. 

The remaining footings apparently give results characteristic of 
the disposition of the steel, indicating inherent inferiority. 

Taking the next group of low values : 

No. 1418. Load eccentric 226 lbs. 

1541. Lack of concrete beloAV bars 231 

1813 223 

1806 241 

No. 1418 and 1541 may be eliminated for assignable causes, but 
the remaining two are apparently normal as far as the record discloses 
the pertinent facts. 

The comparable facts are : 









Load at 




Max. Bond 


No. 


Description 


V 


Failure 


Tension 


at Failure 


1806 


22 % p i. 


0.41 


179,000 


34,800 


241 lbs. 


1813 


12 i/spl.0 


0.39 


121,000 


24,300 


223 



Other dimensional facts being identical, the advantage of larger 
load capacity for small bars is apparent. 



GENERAL INFORMATION 61 

The fact that only one failure occurred in bond with deformed bars 
has already been referred to. The data is as follows : 

Load at Max. Bond 

No. Description p Failure Tension at Failure 

1844 8 %cor. 0.41 269,000 52,000 596 lbs. 

It is obvious that square deformed bars furnish more security in 
bond than equal areas of any other type. 

The distribution of bond in the case of No. 1451 is shown in 
Figure 14. 

This is reported to have failed in bond at the low value of 131 lbs. 
per square inch. 





U ' 


Si 










V 


o 










<s 




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ttll 


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fff 


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4' 














s Ss 












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^ 


1 I 1 ' 


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w 














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rti 


K 


yd 


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p 




rffl 


















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pull 


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n 




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1 


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— §Jy — 


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1 


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8 


8 

s. 

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f * 


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i, 

7" 




^^L-« *_*- 


2' 






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(6 






«l 


t 




Z4-" 


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3' 


3' 


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3" 


3" 


3" 


3-3- 






n. Rein for cinq 


12. It <p Zwcy. Xs 

Typical form of CracK.s-s^^ 
^ Bond failure- // 


































14-51- 





















Fig. 14. Illustrating Increase of Bond Stress Caused by Sloping 
Footing. Failure by Bond. (Talbot.) 



62 HICKS REINFORCED CONCRETE FOOTINGS 

The excess bond caused by sloping the footing is also shown by 
platting the bond line which would exist without sloping, and hatching 
the space between the two lines. 

The results shoAv why footings should net be sloped, unless the 
allowable unit bond stress is reduced proportionately to the reduc- 
tion of jd. 

As the figures for maximum bond and shear at Id do not agree with 
those reported by Talbot, the calculations will be checked. 

No. 1451. p = 0.39. j = 0.89. d = 0.83. 
Load failure = 88,000. 
Soil pressure = 88000/25 = 3520 lbs. 
B/L = 1/5 = .2 — B/L at Id from cap 2.67/5 = .534 
For£/L.534 CKL= (.78) (.10) (5) =.39 
jd at Id from cap = (.89) (7.67) = 6.82 inches 

Shear. Then v = (.39/.83) (120) (.8T5/.89) (8.9/6.82) = 72.4 lbs. 

Talbot determination 59.0 lbs. 
Recheck : 

External shear V at Id from cap = V A W, or 

for B/L .534 = .18 X 88000 = 15840. 
Shear. Then v = 15840/(2.67) (12) (6.82) = 72.5 O. K. 
Steel Stress. For B/L = .2, C KL= (2.84) (.10) (5) =1.42 

Steel required, (.675) (1.8) = 1.21 sq. ins. 

Steel used, 12 l/ 2 " @ 1.96 = 2.35 sq. ins. 

Effective, (2.35) (46/60) =1.80 

Then /, = (1.40/.833) (16000) (1.21/1.80) (.875/89) = 17800 

Talbot determination 17800 

m • p 7 tj (.24) (880 00) i -L - _ 

Magnum Bond. ^= (12)(46/60)(L57) (8 . 9) - = 1641bB. O.K. 

Talbot determination 131 lbs. 

Staggered liars. Designers frequently stagger the lengths of their 
tension steel as the decrease of bending moment permits. 

This should always involve an investigation of the actual bond 
stress available, as this is usually deficient under such circumstances. 

Conclusions. The same principles of design may be applied to 
combined cantilever and raft foundations for depth and steel deter- 
mination. 

It will also be seen that diagrams of the same type for live floor 
loads from zero to 1000 lbs. may be used to design column caps and 
f1;it slabs of the so-called girderless floors. 



